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2x^2-20x+32=0
a = 2; b = -20; c = +32;
Δ = b2-4ac
Δ = -202-4·2·32
Δ = 144
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12}{2*2}=\frac{8}{4} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12}{2*2}=\frac{32}{4} =8 $
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