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(2x+1)(x+4)=10-2x
We move all terms to the left:
(2x+1)(x+4)-(10-2x)=0
We add all the numbers together, and all the variables
(2x+1)(x+4)-(-2x+10)=0
We get rid of parentheses
(2x+1)(x+4)+2x-10=0
We multiply parentheses ..
(+2x^2+8x+x+4)+2x-10=0
We get rid of parentheses
2x^2+8x+x+2x+4-10=0
We add all the numbers together, and all the variables
2x^2+11x-6=0
a = 2; b = 11; c = -6;
Δ = b2-4ac
Δ = 112-4·2·(-6)
Δ = 169
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*2}=\frac{-24}{4} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*2}=\frac{2}{4} =1/2 $
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