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4r^2-2r-90=0
a = 4; b = -2; c = -90;
Δ = b2-4ac
Δ = -22-4·4·(-90)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1444}=38$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-38}{2*4}=\frac{-36}{8} =-4+1/2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+38}{2*4}=\frac{40}{8} =5 $
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