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X^2-32X+248=0
a = 1; b = -32; c = +248;
Δ = b2-4ac
Δ = -322-4·1·248
Δ = 32
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{2}}{2*1}=\frac{32-4\sqrt{2}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{2}}{2*1}=\frac{32+4\sqrt{2}}{2} $
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