5w^2+38w+48=0

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Solution for 5w^2+38w+48=0 equation:



5w^2+38w+48=0
a = 5; b = 38; c = +48;
Δ = b2-4ac
Δ = 382-4·5·48
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-22}{2*5}=\frac{-60}{10} =-6 $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+22}{2*5}=\frac{-16}{10} =-1+3/5 $

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