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4b^2+15b-25=0
a = 4; b = 15; c = -25;
Δ = b2-4ac
Δ = 152-4·4·(-25)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{625}=25$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-25}{2*4}=\frac{-40}{8} =-5 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+25}{2*4}=\frac{10}{8} =1+1/4 $
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