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16x^2-88x+40=0
a = 16; b = -88; c = +40;
Δ = b2-4ac
Δ = -882-4·16·40
Δ = 5184
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{5184}=72$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-88)-72}{2*16}=\frac{16}{32} =1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-88)+72}{2*16}=\frac{160}{32} =5 $
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