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(2x+1)(x+2)=2916
We move all terms to the left:
(2x+1)(x+2)-(2916)=0
We multiply parentheses ..
(+2x^2+4x+x+2)-2916=0
We get rid of parentheses
2x^2+4x+x+2-2916=0
We add all the numbers together, and all the variables
2x^2+5x-2914=0
a = 2; b = 5; c = -2914;
Δ = b2-4ac
Δ = 52-4·2·(-2914)
Δ = 23337
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{23337}=\sqrt{9*2593}=\sqrt{9}*\sqrt{2593}=3\sqrt{2593}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3\sqrt{2593}}{2*2}=\frac{-5-3\sqrt{2593}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3\sqrt{2593}}{2*2}=\frac{-5+3\sqrt{2593}}{4} $
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