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-5b^2+3b+2=0
a = -5; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·(-5)·2
Δ = 49
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{49}=7$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*-5}=\frac{-10}{-10} =1 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*-5}=\frac{4}{-10} =-2/5 $
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