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4x^2-22x-48=0
a = 4; b = -22; c = -48;
Δ = b2-4ac
Δ = -222-4·4·(-48)
Δ = 1252
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1252}=\sqrt{4*313}=\sqrt{4}*\sqrt{313}=2\sqrt{313}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{313}}{2*4}=\frac{22-2\sqrt{313}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{313}}{2*4}=\frac{22+2\sqrt{313}}{8} $
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