4r^2-3r-6=0

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Solution for 4r^2-3r-6=0 equation: 



4r^2-3r-6=0

a = 4; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·4·(-6)
Δ = 105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{105}}{2*4}=\frac{3-\sqrt{105}}{8}
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{105}}{2*4}=\frac{3+\sqrt{105}}{8}
$

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