2x^2+98=180

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Solution for 2x^2+98=180 equation:



2x^2+98=180
We move all terms to the left:
2x^2+98-(180)=0
We add all the numbers together, and all the variables
2x^2-82=0
a = 2; b = 0; c = -82;
Δ = b2-4ac
Δ = 02-4·2·(-82)
Δ = 656
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:
$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{41}}{2*2}=\frac{0-4\sqrt{41}}{4} =-\frac{4\sqrt{41}}{4} =-\sqrt{41} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{41}}{2*2}=\frac{0+4\sqrt{41}}{4} =\frac{4\sqrt{41}}{4} =\sqrt{41} $

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