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7s(3s+1)=4(3+s)
We move all terms to the left:
7s(3s+1)-(4(3+s))=0
We add all the numbers together, and all the variables
7s(3s+1)-(4(s+3))=0
We multiply parentheses
21s^2+7s-(4(s+3))=0
We calculate terms in parentheses: -(4(s+3)), so:We get rid of parentheses
4(s+3)
We multiply parentheses
4s+12
Back to the equation:
-(4s+12)
21s^2+7s-4s-12=0
We add all the numbers together, and all the variables
21s^2+3s-12=0
a = 21; b = 3; c = -12;
Δ = b2-4ac
Δ = 32-4·21·(-12)
Δ = 1017
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1017}=\sqrt{9*113}=\sqrt{9}*\sqrt{113}=3\sqrt{113}$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{113}}{2*21}=\frac{-3-3\sqrt{113}}{42} $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{113}}{2*21}=\frac{-3+3\sqrt{113}}{42} $
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