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10x=3x^2-8
We move all terms to the left:
10x-(3x^2-8)=0
We get rid of parentheses
-3x^2+10x+8=0
a = -3; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·(-3)·8
Δ = 196
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14}{2*-3}=\frac{-24}{-6} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*-3}=\frac{4}{-6} =-2/3 $
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