Solve x^2-x+3+x+3-4=40

$$x ^ { 2 } - x + 3 + x + 3 - 4 = 40$$

Solve for x

$$x=\sqrt{38}\approx 6.164414003$$

$$x=-\sqrt{38}\approx -6.164414003$$

Steps by Finding Square Root

Combine -x and x to get 0.

$$x^{2}+3+3-4=40$$

Add 3 and 3 to get 6.

$$x^{2}+6-4=40$$

Subtract 4 from 6 to get 2.

$$x^{2}+2=40$$

Subtract 2 from both sides.

$$x^{2}=40-2$$

Subtract 2 from 40 to get 38.

$$x^{2}=38$$

Take the square root of both sides of the equation.

$$x=\sqrt{38} x=-\sqrt{38}$$

Steps Using the Quadratic Formula

Combine -x and x to get 0.

$$x^{2}+3+3-4=40$$

Add 3 and 3 to get 6.

$$x^{2}+6-4=40$$

Subtract 4 from 6 to get 2.

$$x^{2}+2=40$$

Subtract 40 from both sides.

$$x^{2}+2-40=0$$

Subtract 40 from 2 to get -38.

$$x^{2}-38=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -38 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-38\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-38\right)}}{2}$$

Multiply -4 times -38.

$$x=\frac{0±\sqrt{152}}{2}$$

Take the square root of 152.

$$x=\frac{0±2\sqrt{38}}{2}$$

Now solve the equation x=\frac{0±2\sqrt{38}}{2} when ± is plus.

$$x=\sqrt{38}$$

Now solve the equation x=\frac{0±2\sqrt{38}}{2} when ± is minus.

$$x=-\sqrt{38}$$

The equation is now solved.

$$x=\sqrt{38} x=-\sqrt{38}$$