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27=w(2w+3)
We move all terms to the left:
27-(w(2w+3))=0
We calculate terms in parentheses: -(w(2w+3)), so:We get rid of parentheses
w(2w+3)
We multiply parentheses
2w^2+3w
Back to the equation:
-(2w^2+3w)
-2w^2-3w+27=0
a = -2; b = -3; c = +27;
Δ = b2-4ac
Δ = -32-4·(-2)·27
Δ = 225
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{225}=15$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-15}{2*-2}=\frac{-12}{-4} =+3 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+15}{2*-2}=\frac{18}{-4} =-4+1/2 $
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