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X^2+8X=38
We move all terms to the left:
X^2+8X-(38)=0
a = 1; b = 8; c = -38;
Δ = b2-4ac
Δ = 82-4·1·(-38)
Δ = 216
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-6\sqrt{6}}{2*1}=\frac{-8-6\sqrt{6}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+6\sqrt{6}}{2*1}=\frac{-8+6\sqrt{6}}{2} $
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