Solve 6t^2=35-t^2

$$6 t ^ { 2 } = 35 - t ^ { 2 }$$

Solve for t

$$t=\sqrt{5}\approx 2.236067977$$

$$t=-\sqrt{5}\approx -2.236067977$$

Steps by Finding Square Root

Add t^{2} to both sides.

$$6t^{2}+t^{2}=35$$

Combine 6t^{2} and t^{2} to get 7t^{2}.

$$7t^{2}=35$$

Divide both sides by 7.

$$t^{2}=\frac{35}{7}$$

Divide 35 by 7 to get 5.

$$t^{2}=5$$

Take the square root of both sides of the equation.

$$t=\sqrt{5} t=-\sqrt{5}$$

Steps Using the Quadratic Formula

Subtract 35 from both sides.

$$6t^{2}-35=-t^{2}$$

Add t^{2} to both sides.

$$6t^{2}-35+t^{2}=0$$

Combine 6t^{2} and t^{2} to get 7t^{2}.

$$7t^{2}-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 0 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{0±\sqrt{0^{2}-4\times 7\left(-35\right)}}{2\times 7}$$

Square 0.

$$t=\frac{0±\sqrt{-4\times 7\left(-35\right)}}{2\times 7}$$

Multiply -4 times 7.

$$t=\frac{0±\sqrt{-28\left(-35\right)}}{2\times 7}$$

Multiply -28 times -35.

$$t=\frac{0±\sqrt{980}}{2\times 7}$$

Take the square root of 980.

$$t=\frac{0±14\sqrt{5}}{2\times 7}$$

Multiply 2 times 7.

$$t=\frac{0±14\sqrt{5}}{14}$$

Now solve the equation t=\frac{0±14\sqrt{5}}{14} when ± is plus.

$$t=\sqrt{5}$$

Now solve the equation t=\frac{0±14\sqrt{5}}{14} when ± is minus.

$$t=-\sqrt{5}$$

The equation is now solved.

$$t=\sqrt{5} t=-\sqrt{5}$$