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x^2+5x=116
We move all terms to the left:
x^2+5x-(116)=0
a = 1; b = 5; c = -116;
Δ = b2-4ac
Δ = 52-4·1·(-116)
Δ = 489
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{-(5)-\sqrt{489}}{2*1}=\frac{-5-\sqrt{489}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{489}}{2*1}=\frac{-5+\sqrt{489}}{2} $
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