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5(2+w)w=10+4(w+1)
We move all terms to the left:
5(2+w)w-(10+4(w+1))=0
We add all the numbers together, and all the variables
5(w+2)w-(10+4(w+1))=0
We multiply parentheses
5w^2+10w-(10+4(w+1))=0
We calculate terms in parentheses: -(10+4(w+1)), so:We get rid of parentheses
10+4(w+1)
determiningTheFunctionDomain 4(w+1)+10
We multiply parentheses
4w+4+10
We add all the numbers together, and all the variables
4w+14
Back to the equation:
-(4w+14)
5w^2+10w-4w-14=0
We add all the numbers together, and all the variables
5w^2+6w-14=0
a = 5; b = 6; c = -14;
Δ = b2-4ac
Δ = 62-4·5·(-14)
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{-(6)-2\sqrt{79}}{2*5}=\frac{-6-2\sqrt{79}}{10} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{79}}{2*5}=\frac{-6+2\sqrt{79}}{10} $
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