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4x^2+3x+1=5
We move all terms to the left:
4x^2+3x+1-(5)=0
We add all the numbers together, and all the variables
4x^2+3x-4=0
a = 4; b = 3; c = -4;
Δ = b2-4ac
Δ = 32-4·4·(-4)
Δ = 73
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{73}}{2*4}=\frac{-3-\sqrt{73}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{73}}{2*4}=\frac{-3+\sqrt{73}}{8} $
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