2x^2+3x=x^2-7x

Solution for 2x^2+3x=x^2-7x equation:

2x^2+3x=x^2-7x

We move all terms to the left:

2x^2+3x-(x^2-7x)=0

We get rid of parentheses

2x^2-x^2+3x+7x=0

We add all the numbers together, and all the variables

x^2+10x=0

a = 1; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·1·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*1}=\frac{-20}{2}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*1}=\frac{0}{2}
=0
$