Solve t ^ 2 + 12 + 36 = 0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$t^{2}+12+36=0$$

Solve for t

$t=-4\sqrt{3}i\approx -0-6.92820323i$
$t=4\sqrt{3}i\approx 6.92820323i$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Add $12$ and $36$ to get $48$.

$$t^{2}+48=0$$

Subtract $48$ from both sides. Anything subtracted from zero gives its negation.

$$t^{2}=-48$$

The equation is now solved.

$$t=4\sqrt{3}i$$ $$t=-4\sqrt{3}i$$

Steps Using the Quadratic Formula

Add $12$ and $36$ to get $48$.

$$t^{2}+48=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $48$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$t=\frac{0±\sqrt{0^{2}-4\times 48}}{2}$$

Square $0$.

$$t=\frac{0±\sqrt{-4\times 48}}{2}$$

Multiply $-4$ times $48$.

$$t=\frac{0±\sqrt{-192}}{2}$$

Take the square root of $-192$.

$$t=\frac{0±8\sqrt{3}i}{2}$$

Now solve the equation $t=\frac{0±8\sqrt{3}i}{2}$ when $±$ is plus.

$$t=4\sqrt{3}i$$

Now solve the equation $t=\frac{0±8\sqrt{3}i}{2}$ when $±$ is minus.

$$t=-4\sqrt{3}i$$

The equation is now solved.

$$t=4\sqrt{3}i$$ $$t=-4\sqrt{3}i$$