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4x(3x-5)=46
We move all terms to the left:
4x(3x-5)-(46)=0
We multiply parentheses
12x^2-20x-46=0
a = 12; b = -20; c = -46;
Δ = b2-4ac
Δ = -202-4·12·(-46)
Δ = 2608
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2608}=\sqrt{16*163}=\sqrt{16}*\sqrt{163}=4\sqrt{163}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{163}}{2*12}=\frac{20-4\sqrt{163}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{163}}{2*12}=\frac{20+4\sqrt{163}}{24} $
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