3x^2+20x-5=2x-5

Solution for 3x^2+20x-5=2x-5 equation:

3x^2+20x-5=2x-5

We move all terms to the left:

3x^2+20x-5-(2x-5)=0

We get rid of parentheses

3x^2+20x-2x+5-5=0

We add all the numbers together, and all the variables

3x^2+18x=0

a = 3; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·3·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*3}=\frac{-36}{6}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*3}=\frac{0}{6}
=0
$