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3x^2+48x+23=0
a = 3; b = 48; c = +23;
Δ = b2-4ac
Δ = 482-4·3·23
Δ = 2028
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2028}=\sqrt{676*3}=\sqrt{676}*\sqrt{3}=26\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-26\sqrt{3}}{2*3}=\frac{-48-26\sqrt{3}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+26\sqrt{3}}{2*3}=\frac{-48+26\sqrt{3}}{6} $
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