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(2w+9)w=95
We move all terms to the left:
(2w+9)w-(95)=0
We multiply parentheses
2w^2+9w-95=0
a = 2; b = 9; c = -95;
Δ = b2-4ac
Δ = 92-4·2·(-95)
Δ = 841
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{841}=29$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-29}{2*2}=\frac{-38}{4} =-9+1/2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+29}{2*2}=\frac{20}{4} =5 $
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