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X^2+7X=116
We move all terms to the left:
X^2+7X-(116)=0
a = 1; b = 7; c = -116;
Δ = b2-4ac
Δ = 72-4·1·(-116)
Δ = 513
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-3\sqrt{57}}{2*1}=\frac{-7-3\sqrt{57}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3\sqrt{57}}{2*1}=\frac{-7+3\sqrt{57}}{2} $
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