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5w(2w-1)=-6w+3
We move all terms to the left:
5w(2w-1)-(-6w+3)=0
We multiply parentheses
10w^2-5w-(-6w+3)=0
We get rid of parentheses
10w^2-5w+6w-3=0
We add all the numbers together, and all the variables
10w^2+w-3=0
a = 10; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·10·(-3)
Δ = 121
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{121}=11$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*10}=\frac{-12}{20} =-3/5 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*10}=\frac{10}{20} =1/2 $
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