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X^2+48X+36=0
a = 1; b = 48; c = +36;
Δ = b2-4ac
Δ = 482-4·1·36
Δ = 2160
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2160}=\sqrt{144*15}=\sqrt{144}*\sqrt{15}=12\sqrt{15}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-12\sqrt{15}}{2*1}=\frac{-48-12\sqrt{15}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+12\sqrt{15}}{2*1}=\frac{-48+12\sqrt{15}}{2} $
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