Solve the quadratic equation 5t 20 6 10t 40 9 7 t

First, the equation is simplified, and the left side can be reduced to [5(t-4) 6] 2+[10(t-4) 9] 2=(25 36)*(t-4) 2+

100 81)*(t-4) 2=(625 324)*(t-4) 2=(7-t) 2 on the right The result is 25*(t-4)=18*+-7-t).

It is easy to get here, t=226 43 or t=-(26 7).

(5t-20)/6]²+10t-40)/9]²=(7-t)²5t-20)^2/36+(10t-40)^2/81=(7-t)^29(5t-20)^2+4(10t-40)^2=324(7-t)^29(25t^2-200t+400)+4(100t^2-800t+1600)=324(49-14t+t^2)

301t^2+464t-5876=0

Apply the formula and do the math yourself.

2x-3y)^2=0

so,2x-3y=0

2x=3yx=3/2y

Any set of x,y, as long as the above equation is satisfied, is the solution of the equation (there are countless sets of modular solutions in this Fang Dan imitation program).

y=x +3x-6 is a quadratic function, not an equation. When y=0, it is a quadratic equation.

Solution of the equation: x=(-3 33) 2

Multiple methods can be used, the first is to use the discriminant formula equal to b-square, -4ac equals 9-4 multiplied by (-6) equals 33, and the second is the matching method.

Considering x+5 as a whole t, the original formula becomes 16t -8t + 1 = 0, and the solution is t = 1 4, so x = -19 4

（x-5）²+2x+t）²=36

x -10x+25+4x +4tx+t =365x +(4t-10)x+(t -11)=0 There are two equal real roots, as determined by the formula.

4t-10）²-4·5·（t²-11）=016t²-80t+100-20t²+220=04t²+80t-320=0

t²+20t-80 =0

t-20)·(t+40）=0

t=20，-40

The root of the equation = -(4t-10) 10

When t=20, x=-7

When t = -40, x = 17

First of all, it's called a quadratic inequality, not an equation!

If you haven't learned about quadratic functions, you can simplify this problem with the methods I've provided.

Let 6t 2-3t-5 = 0

x1,2=（3±√129 ）/12

Since it is required to draw the number line for the part less than or equal to 0, represent x1 and 2 on the number line respectively, use a smooth curve (parabola) through the two points of x1 and x2, and take the lower part of the intersection of the curve and the number line. Of course, there is also method 2 to use the same sign multiplication to get a positive one, and the different sign to multiply a negative one to solve the inequality group separately Then you have to factor it after you find the root.

a=6>0 with the opening pointing upwards and f(t) 0 between the two.

t∈[(3-√129)/12，(3+√129)/12]