-4x^2-4x+12=0

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Solution for -4x^2-4x+12=0 equation:



-4x^2-4x+12=0
a = -4; b = -4; c = +12;
Δ = b2-4ac
Δ = -42-4·(-4)·12
Δ = 208
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:
$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{13}}{2*-4}=\frac{4-4\sqrt{13}}{-8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{13}}{2*-4}=\frac{4+4\sqrt{13}}{-8} $

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