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26-2(2x^2)=4(x+2)+6x
We move all terms to the left:
26-2(2x^2)-(4(x+2)+6x)=0
We calculate terms in parentheses: -(4(x+2)+6x), so:We get rid of parentheses
4(x+2)+6x
We add all the numbers together, and all the variables
6x+4(x+2)
We multiply parentheses
6x+4x+8
We add all the numbers together, and all the variables
10x+8
Back to the equation:
-(10x+8)
-22x^2-10x-8+26=0
We add all the numbers together, and all the variables
-22x^2-10x+18=0
a = -22; b = -10; c = +18;
Δ = b2-4ac
Δ = -102-4·(-22)·18
Δ = 1684
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1684}=\sqrt{4*421}=\sqrt{4}*\sqrt{421}=2\sqrt{421}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{421}}{2*-22}=\frac{10-2\sqrt{421}}{-44} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{421}}{2*-22}=\frac{10+2\sqrt{421}}{-44} $
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