Solve -0.25x^2-1=x^2+3

$$-0.25 { x }^{ 2 } -1= { x }^{ 2 } +3$$

Solve for x (complex solution)

$$x=-\frac{i\times 4\sqrt{5}}{5}\approx -0-1.788854382i$$

$$x=\frac{i\times 4\sqrt{5}}{5}\approx 1.788854382i$$

Steps by Finding Square Root

Subtract x^{2} from both sides.

$$-0.25x^{2}-1-x^{2}=3$$

Combine -0.25x^{2} and -x^{2} to get -1.25x^{2}.

$$-1.25x^{2}-1=3$$

Add 1 to both sides.

$$-1.25x^{2}=3+1$$

Add 3 and 1 to get 4.

$$-1.25x^{2}=4$$

Divide both sides by -1.25.

$$x^{2}=\frac{4}{-1.25}$$

Expand \frac{4}{-1.25} by multiplying both numerator and the denominator by 100.

$$x^{2}=\frac{400}{-125}$$

Reduce the fraction \frac{400}{-125} to lowest terms by extracting and canceling out 25.

$$x^{2}=-\frac{16}{5}$$

The equation is now solved.

$$x=\frac{4\sqrt{5}i}{5} x=-\frac{4\sqrt{5}i}{5}$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$-0.25x^{2}-1-x^{2}=3$$

Combine -0.25x^{2} and -x^{2} to get -1.25x^{2}.

$$-1.25x^{2}-1=3$$

Subtract 3 from both sides.

$$-1.25x^{2}-1-3=0$$

Subtract 3 from -1 to get -4.

$$-1.25x^{2}-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1.25 for a, 0 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-1.25\right)\left(-4\right)}}{2\left(-1.25\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1.25\right)\left(-4\right)}}{2\left(-1.25\right)}$$

Multiply -4 times -1.25.

$$x=\frac{0±\sqrt{5\left(-4\right)}}{2\left(-1.25\right)}$$

Multiply 5 times -4.

$$x=\frac{0±\sqrt{-20}}{2\left(-1.25\right)}$$

Take the square root of -20.

$$x=\frac{0±2\sqrt{5}i}{2\left(-1.25\right)}$$

Multiply 2 times -1.25.

$$x=\frac{0±2\sqrt{5}i}{-2.5}$$

Now solve the equation x=\frac{0±2\sqrt{5}i}{-2.5} when ± is plus.

$$x=-\frac{4\sqrt{5}i}{5}$$

Now solve the equation x=\frac{0±2\sqrt{5}i}{-2.5} when ± is minus.

$$x=\frac{4\sqrt{5}i}{5}$$

The equation is now solved.

$$x=-\frac{4\sqrt{5}i}{5} x=\frac{4\sqrt{5}i}{5}$$