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4x^2-40x+91=0
a = 4; b = -40; c = +91;
Δ = b2-4ac
Δ = -402-4·4·91
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-12}{2*4}=\frac{28}{8} =3+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+12}{2*4}=\frac{52}{8} =6+1/2 $
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