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3+4x^2=43
We move all terms to the left:
3+4x^2-(43)=0
We add all the numbers together, and all the variables
4x^2-40=0
a = 4; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·4·(-40)
Δ = 640
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{10}}{2*4}=\frac{0-8\sqrt{10}}{8} =-\frac{8\sqrt{10}}{8} =-\sqrt{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{10}}{2*4}=\frac{0+8\sqrt{10}}{8} =\frac{8\sqrt{10}}{8} =\sqrt{10} $
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