2x^2+7x+23=x^2-4x

Solution for 2x^2+7x+23=x^2-4x equation:

2x^2+7x+23=x^2-4x

We move all terms to the left:

2x^2+7x+23-(x^2-4x)=0

We get rid of parentheses

2x^2-x^2+7x+4x+23=0

We add all the numbers together, and all the variables

x^2+11x+23=0

a = 1; b = 11; c = +23;
Δ = b2-4ac
Δ = 112-4·1·23
Δ = 29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{29}}{2*1}=\frac{-11-\sqrt{29}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{29}}{2*1}=\frac{-11+\sqrt{29}}{2}
$