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6x^2+19x=36
We move all terms to the left:
6x^2+19x-(36)=0
a = 6; b = 19; c = -36;
Δ = b2-4ac
Δ = 192-4·6·(-36)
Δ = 1225
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-35}{2*6}=\frac{-54}{12} =-4+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+35}{2*6}=\frac{16}{12} =1+1/3 $
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