7r^2+38r+15=0

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Solution for 7r^2+38r+15=0 equation:



7r^2+38r+15=0
a = 7; b = 38; c = +15;
Δ = b2-4ac
Δ = 382-4·7·15
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-32}{2*7}=\frac{-70}{14} =-5 $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+32}{2*7}=\frac{-6}{14} =-3/7 $

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