2r^2-9r+10=0

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Solution for 2r^2-9r+10=0 equation: 



2r^2-9r+10=0

a = 2; b = -9; c = +10;
Δ = b2-4ac
Δ = -92-4·2·10
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-1}{2*2}=\frac{8}{4}
=2
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+1}{2*2}=\frac{10}{4}
=2+1/2
$

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