w^2+4w=75

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Solution for w^2+4w=75 equation:



w^2+4w=75
We move all terms to the left:
w^2+4w-(75)=0
a = 1; b = 4; c = -75;
Δ = b2-4ac
Δ = 42-4·1·(-75)
Δ = 316
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{79}}{2*1}=\frac{-4-2\sqrt{79}}{2} $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{79}}{2*1}=\frac{-4+2\sqrt{79}}{2} $

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