11+2g=2g(g-15)

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Solution for 11+2g=2g(g-15) equation:



11+2g=2g(g-15)
We move all terms to the left:
11+2g-(2g(g-15))=0
We calculate terms in parentheses: -(2g(g-15)), so:
2g(g-15)
We multiply parentheses
2g^2-30g
Back to the equation:
-(2g^2-30g)
We get rid of parentheses
-2g^2+2g+30g+11=0
We add all the numbers together, and all the variables
-2g^2+32g+11=0
a = -2; b = 32; c = +11;
Δ = b2-4ac
Δ = 322-4·(-2)·11
Δ = 1112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{1112}=\sqrt{4*278}=\sqrt{4}*\sqrt{278}=2\sqrt{278}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-2\sqrt{278}}{2*-2}=\frac{-32-2\sqrt{278}}{-4} $
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+2\sqrt{278}}{2*-2}=\frac{-32+2\sqrt{278}}{-4} $

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