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X^2+14X=80
We move all terms to the left:
X^2+14X-(80)=0
a = 1; b = 14; c = -80;
Δ = b2-4ac
Δ = 142-4·1·(-80)
Δ = 516
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{129}}{2*1}=\frac{-14-2\sqrt{129}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{129}}{2*1}=\frac{-14+2\sqrt{129}}{2} $
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