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10x^2-32x-255=0
a = 10; b = -32; c = -255;
Δ = b2-4ac
Δ = -322-4·10·(-255)
Δ = 11224
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11224}=\sqrt{4*2806}=\sqrt{4}*\sqrt{2806}=2\sqrt{2806}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{2806}}{2*10}=\frac{32-2\sqrt{2806}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{2806}}{2*10}=\frac{32+2\sqrt{2806}}{20} $
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