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4a^2-11a-20=0
a = 4; b = -11; c = -20;
Δ = b2-4ac
Δ = -112-4·4·(-20)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{441}=21$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-21}{2*4}=\frac{-10}{8} =-1+1/4 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+21}{2*4}=\frac{32}{8} =4 $
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