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(3x+2)+(x+5)+2x(2x+1)=540
We move all terms to the left:
(3x+2)+(x+5)+2x(2x+1)-(540)=0
We multiply parentheses
4x^2+(3x+2)+(x+5)+2x-540=0
We get rid of parentheses
4x^2+3x+x+2x+2+5-540=0
We add all the numbers together, and all the variables
4x^2+6x-533=0
a = 4; b = 6; c = -533;
Δ = b2-4ac
Δ = 62-4·4·(-533)
Δ = 8564
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8564}=\sqrt{4*2141}=\sqrt{4}*\sqrt{2141}=2\sqrt{2141}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{2141}}{2*4}=\frac{-6-2\sqrt{2141}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{2141}}{2*4}=\frac{-6+2\sqrt{2141}}{8} $
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