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3x^2-22x=-27
We move all terms to the left:
3x^2-22x-(-27)=0
We add all the numbers together, and all the variables
3x^2-22x+27=0
a = 3; b = -22; c = +27;
Δ = b2-4ac
Δ = -222-4·3·27
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-4\sqrt{10}}{2*3}=\frac{22-4\sqrt{10}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+4\sqrt{10}}{2*3}=\frac{22+4\sqrt{10}}{6} $
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