23x^2+40x+15=x^2-2x

Solution for 23x^2+40x+15=x^2-2x equation:

23x^2+40x+15=x^2-2x

We move all terms to the left:

23x^2+40x+15-(x^2-2x)=0

We get rid of parentheses

23x^2-x^2+40x+2x+15=0

We add all the numbers together, and all the variables

22x^2+42x+15=0

a = 22; b = 42; c = +15;
Δ = b2-4ac
Δ = 422-4·22·15
Δ = 444
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:

$\sqrt{\Delta}=\sqrt{444}=\sqrt{4*111}=\sqrt{4}*\sqrt{111}=2\sqrt{111}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{111}}{2*22}=\frac{-42-2\sqrt{111}}{44}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{111}}{2*22}=\frac{-42+2\sqrt{111}}{44}
$