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4x^2+43x+3=0
a = 4; b = 43; c = +3;
Δ = b2-4ac
Δ = 432-4·4·3
Δ = 1801
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{-(43)-\sqrt{1801}}{2*4}=\frac{-43-\sqrt{1801}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+\sqrt{1801}}{2*4}=\frac{-43+\sqrt{1801}}{8} $
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