30x^2+40x=32x^2-40x

Solution for 30x^2+40x=32x^2-40x equation:

30x^2+40x=32x^2-40x

We move all terms to the left:

30x^2+40x-(32x^2-40x)=0

We get rid of parentheses

30x^2-32x^2+40x+40x=0

We add all the numbers together, and all the variables

-2x^2+80x=0

a = -2; b = 80; c = 0;
Δ = b2-4ac
Δ = 802-4·(-2)·0
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-80}{2*-2}=\frac{-160}{-4}
=+40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+80}{2*-2}=\frac{0}{-4}
=0
$