2q2+7q+5=0

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Solution for 2q2+7q+5=0 equation: 



2q^2+7q+5=0

a = 2; b = 7; c = +5;
Δ = b2-4ac
Δ = 72-4·2·5
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-3}{2*2}=\frac{-10}{4}
=-2+1/2
$


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

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