Solve 9d^4-840d^2+631=0

$$9 d ^ { 4 } - 840 d ^ { 2 } + 631 = 0$$

Solve for d

$$d=-\frac{\sqrt{420-3\sqrt{18969}}}{3}\approx -0.870250386$$

$$d=\frac{\sqrt{420-3\sqrt{18969}}}{3}\approx 0.870250386$$

$$d = \frac{\sqrt{3 \sqrt{18969} + 420}}{3} \approx 9.621642147$$

$$d = -\frac{\sqrt{3 \sqrt{18969} + 420}}{3} \approx -9.621642147$$

Steps Using the Quadratic Formula

Substitute t for d^{2}.

$$9t^{2}-840t+631=0$$

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 9 for a, -840 for b, and 631 for c in the quadratic formula.

$$t=\frac{-\left(-840\right)±\sqrt{\left(-840\right)^{2}-4\times 9\times 631}}{2\times 9}$$

Do the calculations.

$$t=\frac{840±6\sqrt{18969}}{18}$$

Solve the equation t=\frac{840±6\sqrt{18969}}{18} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{18969}+140}{3} t=\frac{140-\sqrt{18969}}{3}$$

Since d=t^{2}, the solutions are obtained by evaluating d=±\sqrt{t} for each t.

$$d=\sqrt{\frac{\sqrt{18969}+140}{3}} d=-\sqrt{\frac{\sqrt{18969}+140}{3}} d=\sqrt{\frac{140-\sqrt{18969}}{3}} d=-\sqrt{\frac{140-\sqrt{18969}}{3}}$$