Emergency! Solve a quadratic equation. If you can use factorization, you can use factorization! Than

6y-7)(6y+7)=0

2.(x-3)[4(x-3)-x]=0

x-3)(4x-12-x)=0

3(x-3)(x-4)=0

x²+6x+17=0

6 -17*4*1<0, which cannot be decomposed, so it is unsolvable.

4.(2t+3)[(2t+3)-3]=0

2t(2t+3)=0

2y(y-3)=0

x+√2x-1+√2)x=0

x-8)(x+7)=0

0x-2)(1-x)=0

x-2)(x-1)=0

0x-1)（5x+4)=0

x-9)(x+8)=0

3x²+1=0

3x²>=0

3x²+1>0

No solution. 12.Think of x+1 as a whole.

x+1)-1][(x+1)-2]=0

x(x-1)=0

13.Take advantage of x -y = (x+y)(x-y)x-2) -2x+3) =0

x-2)+(2x+3)][x-2)-(2x+3)]=03x+1)(-x-5)=0

3x+1)(x+5)=0

14.(2x+3)²-4(2x+3)=0

2x+3)(2x+3-4)=0

2x+3)(2x-1)=0

x²-12+27=0

x-3)(x-9)=0

Everything that can be factored is decomposed, and the process is all here.

You don't have to write about the results.

(6y+5)(6y-5)-24=0

36y²+30y-30y-25-24=0

36y²-49=0

6y-7)(6y+7)=0

y=7 6 or y=-7 6

4(x-3)²-x(x-3)=0

x-3)(4x-12-x)=0

x-3)(3x-12)=0

x-3)(x-4)=0

x = 3 or x = 4

x+4)²-2x-1)=0

x²+8x+16-2x+1=0

x²+6x+17=0

6 -17 4 1<0, cannot be decomposed.

Therefore, there is no solution. (2t+3)²=3(2t+3)

2t+3)[(2t+3)-3]=0

2t(2t+3)=0

t=0 or t=-3 2

3-y)²+y²=9

2y²-6y+9-9=0

2y(y-3)=0

y=0 or y=3

1+√2)x²-(1-√2)x=0

x²+√2x²-x+√2x=0

x+√2x-1+√2)x=0

x=0 or x=1-2 2

x²=x+56

x²-x-56=0

x-8)(x+7)=0

x=8 or x=-7

x-2=x(x-2)

x-2-x(x-2)=0

x-2)(1-x)=0

x-2)(x-1)=0

x=2 or x=1

x(5x+4)=5x+4

x(5x+4)-5x+4=0

x-1)（5x+4)=0

x = 1 or x = -4 5

x-1)(x+2)=70

x²-x-2-70=0

x-9)(x+8)=0

x=9 or x=-8

x+1）²+2x(x-1)=0

x²+2x+1+2x²-2x=0

3x²+1=0

Because 3x >=0

So 3x +1>0

Therefore, there is no solution. (x+1)²-3(x+1)+2=0（x+1)-1][(x+1)-2]=0

x(x-1)=0

x=0 or x=1

x-2)²=(2x+3)²

x-2)²-2x+3)²=0

x-2)+(2x+3)][x-2)-(2x+3)]=03x+1)(-x-5)=0

3x+1)(x+5)=0

x=-1 3 or x=-5

2x+3)²=4(2x+3)

2x+3)²-4(2x+3)=0

2x+3)(2x+3-4)=0

2x+3)(2x-1)=0

x = 1 2 or x = -3 2

2(x-3)²=x²-9

2x²-12x+18-x²+9=0

x²-12+27=0

x-3)(x-9)=0

x=3 or x=9

3x+2)(2x-1)=0

Solution: 3x+2=0 or 2x-1=0

x1=-2/3,x2=1/2

2x-1) 2=3 (1-2x).

Solution: 4x 2-4x+1=3-6x

2x^2+x-1=0

x+1)(2x-1)=0

x1=-1, x2=1 Infiltration and Lack of Praise 2

1. The formula for square difference: (2x-1+x)(2x-1-x) = 0(3x-1)(x-1) = 0

3x-1 = 0 or x-1 = 0

The solution is x1 = 1 3, x2 = 1

2、（x-3)²－5-2x)² = 0

x-3)+(5-2x)][x-3)－(5-2x)] = 0(2-x)(3x-8) = 0

2-x= 0 or 3x-8 = 0

x1=2，x2=8/3

x+2)² = (x+2)(x-2)2（x+2)² x+2)(x-2) ＝0(x+2)[2(x+2)-(x-2)] = 0(x+2)(x+6) = 0

x1=-2, x2 = -6

4. Finished, x 11x + 30 = 0(x-5)(x-6) = 0

x1 = 5, x2 = 6.

1: （2x-1）^2-x^2=0

2x-1+x)(2x-1-x)=0 (3x-1)(x-1)=0 x1=1/3 x2=1

2. x^2-6x+9=(5-2x)^2

x-3)^2=(5-2x)^2 (x-3-5+2x)(x-3+5-2x)=0 (3x-8)(-x+2)=0 x1=8/3 x2=2

3. 2(x+2)^2=x^2-4

2(x+2)^2=(x+2)(x-2) 2(x+2)=x-2 x=-6

4. 10(x-3)+x=x^2

x^2-11x+30=0 (x-6)(x-5)=0 x1=5 x2=6

The third question upstairs was done incorrectly.

Please adopt the correct answer in time, and may help you next time, you adopt the correct answer, and you can also get wealth value, thank you.

1 x²-（2a+1）x+a²+a=0

Yes (x-a)(x-(a+1))=0

x = a or a+1

2 . 5x（3x-√2）-3√3 x+√6=015x-5 √2x-3√3 x+√6=0

3x-√2）（5x-√3 ）=0

x= 2 3 or 3 5

3 .(1+√2)x²-(3+√2)x+√2=0.[(1+√2)x-√2](x-1)=0

x = 2 - 2 or 1

4 .x²-3|x|-4=0

x -3x-4 = 0 at 0

x+1）（x-4）=0

When there is x=4 x 0.

i.e. x +3x-4=0

x+4）（x-1）=0

x=-4, so x=4 or -4

(3x-1)^2-（x+2）^2=0

3x-1+x+2）（3x-1-x-2）=0（4x+1）（2x-3）=0

So: 4x+1=0 or 2x-3=0

Solution: x = -1/4 or x = 3/2

The squared difference formula breaks down, and then the equation is solved.