2r^2+r-1.25=0

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Solution for 2r^2+r-1.25=0 equation: 



2r^2+r-1.25=0

a = 2; b = 1; c = -1.25;
Δ = b2-4ac
Δ = 12-4·2·(-1.25)
Δ = 11
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{-(1)-\sqrt{11}}{2*2}=\frac{-1-\sqrt{11}}{4}
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{11}}{2*2}=\frac{-1+\sqrt{11}}{4}
$

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