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x^2-8+2x=15
We move all terms to the left:
x^2-8+2x-(15)=0
We add all the numbers together, and all the variables
x^2+2x-23=0
a = 1; b = 2; c = -23;
Δ = b2-4ac
Δ = 22-4·1·(-23)
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{6}}{2*1}=\frac{-2-4\sqrt{6}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{6}}{2*1}=\frac{-2+4\sqrt{6}}{2} $
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