2b^2+10b+7=0

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Solution for 2b^2+10b+7=0 equation:



2b^2+10b+7=0
a = 2; b = 10; c = +7;
Δ = b2-4ac
Δ = 102-4·2·7
Δ = 44
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{11}}{2*2}=\frac{-10-2\sqrt{11}}{4} $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{11}}{2*2}=\frac{-10+2\sqrt{11}}{4} $

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