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X^2+-16X+40=0
We add all the numbers together, and all the variables
X^2-16X=0
a = 1; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·1·0
Δ = 256
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{256}=16$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*1}=\frac{0}{2} =0 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*1}=\frac{32}{2} =16 $
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