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10p^2+3p-4=0
a = 10; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·10·(-4)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*10}=\frac{-16}{20} =-4/5 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*10}=\frac{10}{20} =1/2 $
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