Solve +5x-(x-2)^2-3(2x+5)=+4(-1)

$$+ 5 x - ( x - 2 ) ^ { 2 } - 3 ( 2 x + 5 ) = + 4 ( - 1 )$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{51}i+3}{2}\approx 1.5-3.570714214i$$

$$x=\frac{3+\sqrt{51}i}{2}\approx 1.5+3.570714214i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$5x-\left(x^{2}-4x+4\right)-3\left(2x+5\right)=4\left(-1\right)$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$5x-x^{2}+4x-4-3\left(2x+5\right)=4\left(-1\right)$$

Combine 5x and 4x to get 9x.

$$9x-x^{2}-4-3\left(2x+5\right)=4\left(-1\right)$$

Use the distributive property to multiply -3 by 2x+5.

$$9x-x^{2}-4-6x-15=4\left(-1\right)$$

Combine 9x and -6x to get 3x.

$$3x-x^{2}-4-15=4\left(-1\right)$$

Subtract 15 from -4 to get -19.

$$3x-x^{2}-19=4\left(-1\right)$$

Multiply 4 and -1 to get -4.

$$3x-x^{2}-19=-4$$

Add 4 to both sides.

$$3x-x^{2}-19+4=0$$

Add -19 and 4 to get -15.

$$3x-x^{2}-15=0$$

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.

$$-x^{2}+3x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4\left(-15\right)}}{2\left(-1\right)}$$

Multiply 4 times -15.

$$x=\frac{-3±\sqrt{9-60}}{2\left(-1\right)}$$

Add 9 to -60.

$$x=\frac{-3±\sqrt{-51}}{2\left(-1\right)}$$

Take the square root of -51.

$$x=\frac{-3±\sqrt{51}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±\sqrt{51}i}{-2}$$

Now solve the equation x=\frac{-3±\sqrt{51}i}{-2} when ± is plus. Add -3 to i\sqrt{51}.

$$x=\frac{-3+\sqrt{51}i}{-2}$$

Divide -3+i\sqrt{51} by -2.

$$x=\frac{-\sqrt{51}i+3}{2}$$

Now solve the equation x=\frac{-3±\sqrt{51}i}{-2} when ± is minus. Subtract i\sqrt{51} from -3.

$$x=\frac{-\sqrt{51}i-3}{-2}$$

Divide -3-i\sqrt{51} by -2.

$$x=\frac{3+\sqrt{51}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{51}i+3}{2} x=\frac{3+\sqrt{51}i}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$5x-\left(x^{2}-4x+4\right)-3\left(2x+5\right)=4\left(-1\right)$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$5x-x^{2}+4x-4-3\left(2x+5\right)=4\left(-1\right)$$

Combine 5x and 4x to get 9x.

$$9x-x^{2}-4-3\left(2x+5\right)=4\left(-1\right)$$

Use the distributive property to multiply -3 by 2x+5.

$$9x-x^{2}-4-6x-15=4\left(-1\right)$$

Combine 9x and -6x to get 3x.

$$3x-x^{2}-4-15=4\left(-1\right)$$

Subtract 15 from -4 to get -19.

$$3x-x^{2}-19=4\left(-1\right)$$

Multiply 4 and -1 to get -4.

$$3x-x^{2}-19=-4$$

Add 19 to both sides.

$$3x-x^{2}=-4+19$$

Add -4 and 19 to get 15.

$$3x-x^{2}=15$$

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.

$$-x^{2}+3x=15$$

Divide both sides by -1.

$$\frac{-x^{2}+3x}{-1}=\frac{15}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{3}{-1}x=\frac{15}{-1}$$

Divide 3 by -1.

$$x^{2}-3x=\frac{15}{-1}$$

Divide 15 by -1.

$$x^{2}-3x=-15$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-15+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-15+\frac{9}{4}$$

Add -15 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=-\frac{51}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=-\frac{51}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{51}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{51}i}{2} x-\frac{3}{2}=-\frac{\sqrt{51}i}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3+\sqrt{51}i}{2} x=\frac{-\sqrt{51}i+3}{2}$$