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6x^2+10=106
We move all terms to the left:
6x^2+10-(106)=0
We add all the numbers together, and all the variables
6x^2-96=0
a = 6; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·6·(-96)
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*6}=\frac{-48}{12} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*6}=\frac{48}{12} =4 $
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