-2r^2+5r+3=0

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Solution for -2r^2+5r+3=0 equation: 



-2r^2+5r+3=0

a = -2; b = 5; c = +3;
Δ = b2-4ac
Δ = 52-4·(-2)·3
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-7}{2*-2}=\frac{-12}{-4}
=+3
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+7}{2*-2}=\frac{2}{-4}
=-1/2
$

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