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22x-3x=6x^2-3x-49
We move all terms to the left:
22x-3x-(6x^2-3x-49)=0
We add all the numbers together, and all the variables
19x-(6x^2-3x-49)=0
We get rid of parentheses
-6x^2+19x+3x+49=0
We add all the numbers together, and all the variables
-6x^2+22x+49=0
a = -6; b = 22; c = +49;
Δ = b2-4ac
Δ = 222-4·(-6)·49
Δ = 1660
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1660}=\sqrt{4*415}=\sqrt{4}*\sqrt{415}=2\sqrt{415}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{415}}{2*-6}=\frac{-22-2\sqrt{415}}{-12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{415}}{2*-6}=\frac{-22+2\sqrt{415}}{-12} $
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