2r^2+3r-4=0

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



2r^2+3r-4=0

a = 2; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·2·(-4)
Δ = 41
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{41}}{2*2}=\frac{-3-\sqrt{41}}{4}
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{41}}{2*2}=\frac{-3+\sqrt{41}}{4}
$

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