5b^2+14b-12=0

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Solution for 5b^2+14b-12=0 equation:



5b^2+14b-12=0
a = 5; b = 14; c = -12;
Δ = b2-4ac
Δ = 142-4·5·(-12)
Δ = 436
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:
$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{109}}{2*5}=\frac{-14-2\sqrt{109}}{10} $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{109}}{2*5}=\frac{-14+2\sqrt{109}}{10} $

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