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3x(x-5)=48
We move all terms to the left:
3x(x-5)-(48)=0
We multiply parentheses
3x^2-15x-48=0
a = 3; b = -15; c = -48;
Δ = b2-4ac
Δ = -152-4·3·(-48)
Δ = 801
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{89}}{2*3}=\frac{15-3\sqrt{89}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{89}}{2*3}=\frac{15+3\sqrt{89}}{6} $
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