Solve q^6-28q^3+27=0

$$q ^ { 6 } - 28 q ^ { 3 } + 27 = 0$$

Solve for q

$$q=1$$

$$q=3$$

Steps Using the Quadratic Formula

Substitute t for q^{3}.

$$t^{2}-28t+27=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -28 for b, and 27 for c in the quadratic formula.

$$t=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 1\times 27}}{2}$$

Do the calculations.

$$t=\frac{28±26}{2}$$

Solve the equation t=\frac{28±26}{2} when ± is plus and when ± is minus.

$$t=27 t=1$$

Since q=t^{3}, the solutions are obtained by evaluating q=\sqrt[3]{t} for each t.

$$q=3 q=1$$