Solve 7=2+20x-5x^2

$$7=2+20x-5 { x }^{ 2 }$$

Solve for x

$$x=\sqrt{3}+2\approx 3.732050808$$

$$x=2-\sqrt{3}\approx 0.267949192$$

Steps Using the Quadratic Formula

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

$$2+20x-5x^{2}=7$$

Subtract 7 from both sides.

$$2+20x-5x^{2}-7=0$$

Subtract 7 from 2 to get -5.

$$-5+20x-5x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-5x^{2}+20x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 20 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-5\right)\left(-5\right)}}{2\left(-5\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-5\right)\left(-5\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-20±\sqrt{400+20\left(-5\right)}}{2\left(-5\right)}$$

Multiply 20 times -5.

$$x=\frac{-20±\sqrt{400-100}}{2\left(-5\right)}$$

Add 400 to -100.

$$x=\frac{-20±\sqrt{300}}{2\left(-5\right)}$$

Take the square root of 300.

$$x=\frac{-20±10\sqrt{3}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-20±10\sqrt{3}}{-10}$$

Now solve the equation x=\frac{-20±10\sqrt{3}}{-10} when ± is plus. Add -20 to 10\sqrt{3}.

$$x=\frac{10\sqrt{3}-20}{-10}$$

Divide -20+10\sqrt{3} by -10.

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

Now solve the equation x=\frac{-20±10\sqrt{3}}{-10} when ± is minus. Subtract 10\sqrt{3} from -20.

$$x=\frac{-10\sqrt{3}-20}{-10}$$

Divide -20-10\sqrt{3} by -10.

$$x=\sqrt{3}+2$$

The equation is now solved.

$$x=2-\sqrt{3} x=\sqrt{3}+2$$

Steps for Completing the Square

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

$$2+20x-5x^{2}=7$$

Subtract 2 from both sides.

$$20x-5x^{2}=7-2$$

Subtract 2 from 7 to get 5.

$$20x-5x^{2}=5$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-5x^{2}+20x=5$$

Divide both sides by -5.

$$\frac{-5x^{2}+20x}{-5}=\frac{5}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{20}{-5}x=\frac{5}{-5}$$

Divide 20 by -5.

$$x^{2}-4x=\frac{5}{-5}$$

Divide 5 by -5.

$$x^{2}-4x=-1$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-4x+\left(-2\right)^{2}=-1+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-1+4$$

Add -1 to 4.

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

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{3}$$

Simplify.

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

Add 2 to both sides of the equation.

$$x=\sqrt{3}+2 x=2-\sqrt{3}$$