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20-15a-5a^2=0
a = -5; b = -15; c = +20;
Δ = b2-4ac
Δ = -152-4·(-5)·20
Δ = 625
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{625}=25$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-25}{2*-5}=\frac{-10}{-10} =1 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+25}{2*-5}=\frac{40}{-10} =-4 $
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