7x^2-43x+40=5x^2-79X+16

Solution for 7x^2-43x+40=5x^2-79X+16 equation:

7x^2-43x+40=5x^2-79x+16

We move all terms to the left:

7x^2-43x+40-(5x^2-79x+16)=0

We get rid of parentheses

7x^2-5x^2-43x+79x-16+40=0

We add all the numbers together, and all the variables

2x^2+36x+24=0

a = 2; b = 36; c = +24;
Δ = b2-4ac
Δ = 362-4·2·24
Δ = 1104
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1104}=\sqrt{16*69}=\sqrt{16}*\sqrt{69}=4\sqrt{69}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{69}}{2*2}=\frac{-36-4\sqrt{69}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{69}}{2*2}=\frac{-36+4\sqrt{69}}{4}
$