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6x^2+10=256
We move all terms to the left:
6x^2+10-(256)=0
We add all the numbers together, and all the variables
6x^2-246=0
a = 6; b = 0; c = -246;
Δ = b2-4ac
Δ = 02-4·6·(-246)
Δ = 5904
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5904}=\sqrt{144*41}=\sqrt{144}*\sqrt{41}=12\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{41}}{2*6}=\frac{0-12\sqrt{41}}{12} =-\frac{12\sqrt{41}}{12} =-\sqrt{41} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{41}}{2*6}=\frac{0+12\sqrt{41}}{12} =\frac{12\sqrt{41}}{12} =\sqrt{41} $
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