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2x^2+9x=68
We move all terms to the left:
2x^2+9x-(68)=0
a = 2; b = 9; c = -68;
Δ = b2-4ac
Δ = 92-4·2·(-68)
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-25}{2*2}=\frac{-34}{4} =-8+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+25}{2*2}=\frac{16}{4} =4 $
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