x^2+28x-23=40

Solution for x^2+28x-23=40 equation:

x^2+28x-23=40

We move all terms to the left:

x^2+28x-23-(40)=0

We add all the numbers together, and all the variables

x^2+28x-63=0

a = 1; b = 28; c = -63;
Δ = b2-4ac
Δ = 282-4·1·(-63)
Δ = 1036
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:

$\sqrt{\Delta}=\sqrt{1036}=\sqrt{4*259}=\sqrt{4}*\sqrt{259}=2\sqrt{259}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{259}}{2*1}=\frac{-28-2\sqrt{259}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{259}}{2*1}=\frac{-28+2\sqrt{259}}{2}
$