2x^2=7-4x=-2x^2-4x+7=0

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Solution for 2x^2=7-4x=-2x^2-4x+7=0 equation:



2x^2=7-4x=-2x^2-4x+7=0
We move all terms to the left:
2x^2-(7-4x)=0
We add all the numbers together, and all the variables
2x^2-(-4x+7)=0
We get rid of parentheses
2x^2+4x-7=0
a = 2; b = 4; c = -7;
Δ = b2-4ac
Δ = 42-4·2·(-7)
Δ = 72
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:
$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{2}}{2*2}=\frac{-4-6\sqrt{2}}{4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{2}}{2*2}=\frac{-4+6\sqrt{2}}{4} $

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