Solve x ^ 2 + 4 ^ 2 * c - 12 = 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

$$x^{2}+4^{2}c-12=0$$

Solve for c

$c=-\frac{x^{2}}{16}+\frac{3}{4}$

Steps for Solving Linear Equation

Calculate $4$ to the power of $2$ and get $16$.

$$x^{2}+16c-12=0$$

Subtract $x^{2}$ from both sides. Anything subtracted from zero gives its negation.

$$16c-12=-x^{2}$$

Add $12$ to both sides.

$$16c=-x^{2}+12$$

The equation is in standard form.

$$16c=12-x^{2}$$

Divide both sides by $16$.

$$\frac{16c}{16}=\frac{12-x^{2}}{16}$$

Dividing by $16$ undoes the multiplication by $16$.

$$c=\frac{12-x^{2}}{16}$$

Divide $-x^{2}+12$ by $16$.

$$c=-\frac{x^{2}}{16}+\frac{3}{4}$$

Solve for x (complex solution)

$x=-2\sqrt{3-4c}$
$x=2\sqrt{3-4c}$

Solve for x

$x=2\sqrt{3-4c}$
$x=-2\sqrt{3-4c}\text{, }c\leq \frac{3}{4}$