Solve {l}{-x^2+5-1=2x*2x-3}{text{B)}}

$$\left. \begin{array} { l } { - x ^ { 2 } + 5 - 1 = 2 x \cdot 2 x - 3 } \\ { \text { B) } } \end{array} \right.$$

Solve for x

$$x = \frac{\sqrt{35}}{5} \approx 1.183215957$$

$$x = -\frac{\sqrt{35}}{5} \approx -1.183215957$$

Steps by Finding Square Root

Multiply x and x to get x^{2}.

$$-x^{2}+5-1=2x^{2}\times 2-3$$

Subtract 1 from 5 to get 4.

$$-x^{2}+4=2x^{2}\times 2-3$$

Multiply 2 and 2 to get 4.

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

Subtract 4x^{2} from both sides.

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

Subtract 4 from both sides.

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

Subtract 4 from -3 to get -7.

$$-x^{2}-4x^{2}=-7$$

Combine -x^{2} and -4x^{2} to get -5x^{2}.

$$-5x^{2}=-7$$

Divide both sides by -5.

$$x^{2}=\frac{-7}{-5}$$

Fraction \frac{-7}{-5} can be simplified to \frac{7}{5} by removing the negative sign from both the numerator and the denominator.

$$x^{2}=\frac{7}{5}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{35}}{5} x=-\frac{\sqrt{35}}{5}$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$-x^{2}+5-1=2x^{2}\times 2-3$$

Subtract 1 from 5 to get 4.

$$-x^{2}+4=2x^{2}\times 2-3$$

Multiply 2 and 2 to get 4.

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

Subtract 4x^{2} from both sides.

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

Add 3 to both sides.

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

Add 4 and 3 to get 7.

$$-x^{2}+7-4x^{2}=0$$

Combine -x^{2} and -4x^{2} to get -5x^{2}.

$$-5x^{2}+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 0 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-5\right)\times 7}}{2\left(-5\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-5\right)\times 7}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{0±\sqrt{20\times 7}}{2\left(-5\right)}$$

Multiply 20 times 7.

$$x=\frac{0±\sqrt{140}}{2\left(-5\right)}$$

Take the square root of 140.

$$x=\frac{0±2\sqrt{35}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{0±2\sqrt{35}}{-10}$$

Now solve the equation x=\frac{0±2\sqrt{35}}{-10} when ± is plus.

$$x=-\frac{\sqrt{35}}{5}$$

Now solve the equation x=\frac{0±2\sqrt{35}}{-10} when ± is minus.

$$x=\frac{\sqrt{35}}{5}$$

The equation is now solved.

$$x=-\frac{\sqrt{35}}{5} x=\frac{\sqrt{35}}{5}$$