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6x(4x-5)=13
We move all terms to the left:
6x(4x-5)-(13)=0
We multiply parentheses
24x^2-30x-13=0
a = 24; b = -30; c = -13;
Δ = b2-4ac
Δ = -302-4·24·(-13)
Δ = 2148
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2148}=\sqrt{4*537}=\sqrt{4}*\sqrt{537}=2\sqrt{537}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{537}}{2*24}=\frac{30-2\sqrt{537}}{48} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{537}}{2*24}=\frac{30+2\sqrt{537}}{48} $
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