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x2-20x+96=0
We add all the numbers together, and all the variables
x^2-20x+96=0
a = 1; b = -20; c = +96;
Δ = b2-4ac
Δ = -202-4·1·96
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*1}=\frac{16}{2} =8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*1}=\frac{24}{2} =12 $
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