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3a-5+8a^2=0
a = 8; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·8·(-5)
Δ = 169
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{169}=13$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*8}=\frac{-16}{16} =-1 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*8}=\frac{10}{16} =5/8 $
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