Solve 4 - 8 = 3x ^ 2 + 5 | 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

$$4-8=3x^{2}+5$$

Solve for x (complex solution)

$x=-\sqrt{3}i\approx -0-1.732050808i$
$x=\sqrt{3}i\approx 1.732050808i$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Subtract $8$ from $4$ to get $-4$.

$$-4=3x^{2}+5$$

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}+5=-4$$

Subtract $5$ from both sides.

$$3x^{2}=-4-5$$

Subtract $5$ from $-4$ to get $-9$.

$$3x^{2}=-9$$

Divide both sides by $3$.

$$x^{2}=\frac{-9}{3}$$

Divide $-9$ by $3$ to get $-3$.

$$x^{2}=-3$$

The equation is now solved.

$$x=\sqrt{3}i$$ $$x=-\sqrt{3}i$$

Steps Using the Quadratic Formula

Subtract $8$ from $4$ to get $-4$.

$$-4=3x^{2}+5$$

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}+5=-4$$

Add $4$ to both sides.

$$3x^{2}+5+4=0$$

Add $5$ and $4$ to get $9$.

$$3x^{2}+9=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $0$ for $b$, and $9$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\times 9}}{2\times 3}$$

Square $0$.

$$x=\frac{0±\sqrt{-4\times 3\times 9}}{2\times 3}$$

Multiply $-4$ times $3$.

$$x=\frac{0±\sqrt{-12\times 9}}{2\times 3}$$

Multiply $-12$ times $9$.

$$x=\frac{0±\sqrt{-108}}{2\times 3}$$

Take the square root of $-108$.

$$x=\frac{0±6\sqrt{3}i}{2\times 3}$$

Multiply $2$ times $3$.

$$x=\frac{0±6\sqrt{3}i}{6}$$

Now solve the equation $x=\frac{0±6\sqrt{3}i}{6}$ when $±$ is plus.

$$x=\sqrt{3}i$$

Now solve the equation $x=\frac{0±6\sqrt{3}i}{6}$ when $±$ is minus.

$$x=-\sqrt{3}i$$

The equation is now solved.

$$x=\sqrt{3}i$$ $$x=-\sqrt{3}i$$