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(2x+10)(x+6)=258+x
We move all terms to the left:
(2x+10)(x+6)-(258+x)=0
We add all the numbers together, and all the variables
(2x+10)(x+6)-(x+258)=0
We get rid of parentheses
(2x+10)(x+6)-x-258=0
We multiply parentheses ..
(+2x^2+12x+10x+60)-x-258=0
We add all the numbers together, and all the variables
(+2x^2+12x+10x+60)-1x-258=0
We get rid of parentheses
2x^2+12x+10x-1x+60-258=0
We add all the numbers together, and all the variables
2x^2+21x-198=0
a = 2; b = 21; c = -198;
Δ = b2-4ac
Δ = 212-4·2·(-198)
Δ = 2025
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{2025}=45$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-45}{2*2}=\frac{-66}{4} =-16+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+45}{2*2}=\frac{24}{4} =6 $
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