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6x^2-4x-28=-10+5x
We move all terms to the left:
6x^2-4x-28-(-10+5x)=0
We add all the numbers together, and all the variables
6x^2-4x-(5x-10)-28=0
We get rid of parentheses
6x^2-4x-5x+10-28=0
We add all the numbers together, and all the variables
6x^2-9x-18=0
a = 6; b = -9; c = -18;
Δ = b2-4ac
Δ = -92-4·6·(-18)
Δ = 513
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{57}}{2*6}=\frac{9-3\sqrt{57}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{57}}{2*6}=\frac{9+3\sqrt{57}}{12} $
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