-16x^2+5x+20=0

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Solution for -16x^2+5x+20=0 equation:



-16x^2+5x+20=0
a = -16; b = 5; c = +20;
Δ = b2-4ac
Δ = 52-4·(-16)·20
Δ = 1305
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:
$\sqrt{\Delta}=\sqrt{1305}=\sqrt{9*145}=\sqrt{9}*\sqrt{145}=3\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3\sqrt{145}}{2*-16}=\frac{-5-3\sqrt{145}}{-32} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3\sqrt{145}}{2*-16}=\frac{-5+3\sqrt{145}}{-32} $

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