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2x^2+3=245
We move all terms to the left:
2x^2+3-(245)=0
We add all the numbers together, and all the variables
2x^2-242=0
a = 2; b = 0; c = -242;
Δ = b2-4ac
Δ = 02-4·2·(-242)
Δ = 1936
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{1936}=44$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-44}{2*2}=\frac{-44}{4} =-11 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+44}{2*2}=\frac{44}{4} =11 $
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