7b2+10b+3=0

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Solution for 7b2+10b+3=0 equation: 



7b^2+10b+3=0

a = 7; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·7·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4}{2*7}=\frac{-14}{14}
=-1
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4}{2*7}=\frac{-6}{14}
=-3/7
$

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