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12x^2-10=38
We move all terms to the left:
12x^2-10-(38)=0
We add all the numbers together, and all the variables
12x^2-48=0
a = 12; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·12·(-48)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*12}=\frac{-48}{24} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*12}=\frac{48}{24} =2 $
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