-4x^2+18x-16=0

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Solution for -4x^2+18x-16=0 equation:



-4x^2+18x-16=0
a = -4; b = 18; c = -16;
Δ = b2-4ac
Δ = 182-4·(-4)·(-16)
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{17}}{2*-4}=\frac{-18-2\sqrt{17}}{-8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{17}}{2*-4}=\frac{-18+2\sqrt{17}}{-8} $

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