2b2-3b-7=0

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Solution for 2b2-3b-7=0 equation:



2b^2-3b-7=0
a = 2; b = -3; c = -7;
Δ = b2-4ac
Δ = -32-4·2·(-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{-(-3)-\sqrt{65}}{2*2}=\frac{3-\sqrt{65}}{4} $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{65}}{2*2}=\frac{3+\sqrt{65}}{4} $

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