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4x^2-35x+4=0
a = 4; b = -35; c = +4;
Δ = b2-4ac
Δ = -352-4·4·4
Δ = 1161
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1161}=\sqrt{9*129}=\sqrt{9}*\sqrt{129}=3\sqrt{129}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-3\sqrt{129}}{2*4}=\frac{35-3\sqrt{129}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+3\sqrt{129}}{2*4}=\frac{35+3\sqrt{129}}{8} $
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