2x^2+7x-16=0

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Solution for 2x^2+7x-16=0 equation: 



2x^2+7x-16=0

a = 2; b = 7; c = -16;
Δ = b2-4ac
Δ = 72-4·2·(-16)
Δ = 177
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{177}}{2*2}=\frac{-7-\sqrt{177}}{4}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{177}}{2*2}=\frac{-7+\sqrt{177}}{4}
$

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