4x^2+16x-50=0

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Solution for 4x^2+16x-50=0 equation:



4x^2+16x-50=0
a = 4; b = 16; c = -50;
Δ = b2-4ac
Δ = 162-4·4·(-50)
Δ = 1056
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:
$\sqrt{\Delta}=\sqrt{1056}=\sqrt{16*66}=\sqrt{16}*\sqrt{66}=4\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{66}}{2*4}=\frac{-16-4\sqrt{66}}{8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{66}}{2*4}=\frac{-16+4\sqrt{66}}{8} $

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