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3x^2=10x+4
We move all terms to the left:
3x^2-(10x+4)=0
We get rid of parentheses
3x^2-10x-4=0
a = 3; b = -10; c = -4;
Δ = b2-4ac
Δ = -102-4·3·(-4)
Δ = 148
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{37}}{2*3}=\frac{10-2\sqrt{37}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{37}}{2*3}=\frac{10+2\sqrt{37}}{6} $
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