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(2x+3)(x+1)-5(4x+1)=(2x-3.5)(x+1)5
We move all terms to the left:
(2x+3)(x+1)-5(4x+1)-((2x-3.5)(x+1)5)=0
We multiply parentheses
(2x+3)(x+1)-20x-((2x-3.5)(x+1)5)-5=0
We multiply parentheses ..
(+2x^2+2x+3x+3)-20x-((2x-3.5)(x+1)5)-5=0
We calculate terms in parentheses: -((2x-3.5)(x+1)5), so:We get rid of parentheses
(2x-3.5)(x+1)5
We multiply parentheses ..
(+2x^2+2x-3.5x-3.5)5
We multiply parentheses
10x^2+10x-15x-17.5
We add all the numbers together, and all the variables
10x^2-5x-17.5
Back to the equation:
-(10x^2-5x-17.5)
2x^2-10x^2+2x+3x-20x+5x+3+17.5-5=0
We add all the numbers together, and all the variables
-8x^2-10x+15.5=0
a = -8; b = -10; c = +15.5;
Δ = b2-4ac
Δ = -102-4·(-8)·15.5
Δ = 596
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{596}=\sqrt{4*149}=\sqrt{4}*\sqrt{149}=2\sqrt{149}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{149}}{2*-8}=\frac{10-2\sqrt{149}}{-16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{149}}{2*-8}=\frac{10+2\sqrt{149}}{-16} $
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