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2x^2-36x+124=0
a = 2; b = -36; c = +124;
Δ = b2-4ac
Δ = -362-4·2·124
Δ = 304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{19}}{2*2}=\frac{36-4\sqrt{19}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{19}}{2*2}=\frac{36+4\sqrt{19}}{4} $
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