2b^2+11b+7=0

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Solution for 2b^2+11b+7=0 equation: 



2b^2+11b+7=0

a = 2; b = 11; c = +7;
Δ = b2-4ac
Δ = 112-4·2·7
Δ = 65
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{-(11)-\sqrt{65}}{2*2}=\frac{-11-\sqrt{65}}{4}
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{65}}{2*2}=\frac{-11+\sqrt{65}}{4}
$

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