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(2x+6)(4x+7)=(6x+13)
We move all terms to the left:
(2x+6)(4x+7)-((6x+13))=0
We multiply parentheses ..
(+8x^2+14x+24x+42)-((6x+13))=0
We calculate terms in parentheses: -((6x+13)), so:We get rid of parentheses
(6x+13)
We get rid of parentheses
6x+13
Back to the equation:
-(6x+13)
8x^2+14x+24x-6x+42-13=0
We add all the numbers together, and all the variables
8x^2+32x+29=0
a = 8; b = 32; c = +29;
Δ = b2-4ac
Δ = 322-4·8·29
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:x_{1}=\frac{-b-\sqrt{\Delta}}{2a}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}
The end solution:
\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{6}}{2*8}=\frac{-32-4\sqrt{6}}{16}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{6}}{2*8}=\frac{-32+4\sqrt{6}}{16}
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