7p^2+11p+4=0

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Solution for 7p^2+11p+4=0 equation: 



7p^2+11p+4=0

a = 7; b = 11; c = +4;
Δ = b2-4ac
Δ = 112-4·7·4
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*7}=\frac{-14}{14}
=-1
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*7}=\frac{-8}{14}
=-4/7
$

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