5b2+3b-4=0

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Solution for 5b2+3b-4=0 equation: 



5b^2+3b-4=0

a = 5; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·5·(-4)
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{89}}{2*5}=\frac{-3-\sqrt{89}}{10}
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{89}}{2*5}=\frac{-3+\sqrt{89}}{10}
$

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