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x(4x+1)=68
We move all terms to the left:
x(4x+1)-(68)=0
We multiply parentheses
4x^2+x-68=0
a = 4; b = 1; c = -68;
Δ = b2-4ac
Δ = 12-4·4·(-68)
Δ = 1089
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{1089}=33$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-33}{2*4}=\frac{-34}{8} =-4+1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+33}{2*4}=\frac{32}{8} =4 $
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