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2x^2+11=4x+49
We move all terms to the left:
2x^2+11-(4x+49)=0
We get rid of parentheses
2x^2-4x-49+11=0
We add all the numbers together, and all the variables
2x^2-4x-38=0
a = 2; b = -4; c = -38;
Δ = b2-4ac
Δ = -42-4·2·(-38)
Δ = 320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{5}}{2*2}=\frac{4-8\sqrt{5}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{5}}{2*2}=\frac{4+8\sqrt{5}}{4} $
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