-5x^2+40x=-7

Solution for -5x^2+40x=-7 equation:

-5x^2+40x=-7

We move all terms to the left:

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

We add all the numbers together, and all the variables

-5x^2+40x+7=0

a = -5; b = 40; c = +7;
Δ = b2-4ac
Δ = 402-4·(-5)·7
Δ = 1740
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:

$\sqrt{\Delta}=\sqrt{1740}=\sqrt{4*435}=\sqrt{4}*\sqrt{435}=2\sqrt{435}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{435}}{2*-5}=\frac{-40-2\sqrt{435}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{435}}{2*-5}=\frac{-40+2\sqrt{435}}{-10}
$