7w+w^2+12=0

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Solution for 7w+w^2+12=0 equation: 



7w+w^2+12=0

a = 1; b = 7; c = +12;
Δ = b2-4ac
Δ = 72-4·1·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*1}=\frac{-8}{2}
=-4
$


$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*1}=\frac{-6}{2}
=-3
$

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