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4x^2+28x+43=0
a = 4; b = 28; c = +43;
Δ = b2-4ac
Δ = 282-4·4·43
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{6}}{2*4}=\frac{-28-4\sqrt{6}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{6}}{2*4}=\frac{-28+4\sqrt{6}}{8} $
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