-4w^2+23w-6=0

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Solution for -4w^2+23w-6=0 equation: 



-4w^2+23w-6=0

a = -4; b = 23; c = -6;
Δ = b2-4ac
Δ = 232-4·(-4)·(-6)
Δ = 433
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{433}}{2*-4}=\frac{-23-\sqrt{433}}{-8}
$


$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{433}}{2*-4}=\frac{-23+\sqrt{433}}{-8}
$

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