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(x-3)(2x+1)=4x+3x+13
We move all terms to the left:
(x-3)(2x+1)-(4x+3x+13)=0
We add all the numbers together, and all the variables
(x-3)(2x+1)-(7x+13)=0
We get rid of parentheses
(x-3)(2x+1)-7x-13=0
We multiply parentheses ..
(+2x^2+x-6x-3)-7x-13=0
We get rid of parentheses
2x^2+x-6x-7x-3-13=0
We add all the numbers together, and all the variables
2x^2-12x-16=0
a = 2; b = -12; c = -16;
Δ = b2-4ac
Δ = -122-4·2·(-16)
Δ = 272
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{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{17}}{2*2}=\frac{12-4\sqrt{17}}{4}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{17}}{2*2}=\frac{12+4\sqrt{17}}{4}
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