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3x-4=2-5x^2
We move all terms to the left:
3x-4-(2-5x^2)=0
We get rid of parentheses
5x^2+3x-2-4=0
We add all the numbers together, and all the variables
5x^2+3x-6=0
a = 5; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·5·(-6)
Δ = 129
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{129}}{2*5}=\frac{-3-\sqrt{129}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{129}}{2*5}=\frac{-3+\sqrt{129}}{10} $
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