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4x^2-14x+3=0
a = 4; b = -14; c = +3;
Δ = b2-4ac
Δ = -142-4·4·3
Δ = 148
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{37}}{2*4}=\frac{14-2\sqrt{37}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{37}}{2*4}=\frac{14+2\sqrt{37}}{8} $
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