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3x^2+4x+.75=0
We add all the numbers together, and all the variables
3x^2+4x+0.75=0
a = 3; b = 4; c = +0.75;
Δ = b2-4ac
Δ = 42-4·3·0.75
Δ = 7
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{-(4)-\sqrt{7}}{2*3}=\frac{-4-\sqrt{7}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{7}}{2*3}=\frac{-4+\sqrt{7}}{6} $
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