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6x(2x+5)=11
We move all terms to the left:
6x(2x+5)-(11)=0
We multiply parentheses
12x^2+30x-11=0
a = 12; b = 30; c = -11;
Δ = b2-4ac
Δ = 302-4·12·(-11)
Δ = 1428
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1428}=\sqrt{4*357}=\sqrt{4}*\sqrt{357}=2\sqrt{357}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{357}}{2*12}=\frac{-30-2\sqrt{357}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{357}}{2*12}=\frac{-30+2\sqrt{357}}{24} $
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