Solve (3x+(1+2x)(1-2x)=61x+50(x+3)

$$( 3 x + ( 1 + 2 x ) ( 1 - 2 x ) = 61 x + 50 ( x + 3 )$$

Solve for x

$$x=\sqrt{145}-\frac{27}{2}\approx -1.458405421$$

$$x=-\sqrt{145}-\frac{27}{2}\approx -25.541594579$$

Steps Using the Quadratic Formula

Consider \left(1+2x\right)\left(1-2x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$3x+1-\left(2x\right)^{2}=61x+50\left(x+3\right)$$

Expand \left(2x\right)^{2}.

$$3x+1-2^{2}x^{2}=61x+50\left(x+3\right)$$

Calculate 2 to the power of 2 and get 4.

$$3x+1-4x^{2}=61x+50\left(x+3\right)$$

Use the distributive property to multiply 50 by x+3.

$$3x+1-4x^{2}=61x+50x+150$$

Combine 61x and 50x to get 111x.

$$3x+1-4x^{2}=111x+150$$

Subtract 111x from both sides.

$$3x+1-4x^{2}-111x=150$$

Combine 3x and -111x to get -108x.

$$-108x+1-4x^{2}=150$$

Subtract 150 from both sides.

$$-108x+1-4x^{2}-150=0$$

Subtract 150 from 1 to get -149.

$$-108x-149-4x^{2}=0$$

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.

$$-4x^{2}-108x-149=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -108 for b, and -149 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-108\right)±\sqrt{\left(-108\right)^{2}-4\left(-4\right)\left(-149\right)}}{2\left(-4\right)}$$

Square -108.

$$x=\frac{-\left(-108\right)±\sqrt{11664-4\left(-4\right)\left(-149\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-108\right)±\sqrt{11664+16\left(-149\right)}}{2\left(-4\right)}$$

Multiply 16 times -149.

$$x=\frac{-\left(-108\right)±\sqrt{11664-2384}}{2\left(-4\right)}$$

Add 11664 to -2384.

$$x=\frac{-\left(-108\right)±\sqrt{9280}}{2\left(-4\right)}$$

Take the square root of 9280.

$$x=\frac{-\left(-108\right)±8\sqrt{145}}{2\left(-4\right)}$$

The opposite of -108 is 108.

$$x=\frac{108±8\sqrt{145}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{108±8\sqrt{145}}{-8}$$

Now solve the equation x=\frac{108±8\sqrt{145}}{-8} when ± is plus. Add 108 to 8\sqrt{145}.

$$x=\frac{8\sqrt{145}+108}{-8}$$

Divide 108+8\sqrt{145} by -8.

$$x=-\sqrt{145}-\frac{27}{2}$$

Now solve the equation x=\frac{108±8\sqrt{145}}{-8} when ± is minus. Subtract 8\sqrt{145} from 108.

$$x=\frac{108-8\sqrt{145}}{-8}$$

Divide 108-8\sqrt{145} by -8.

$$x=\sqrt{145}-\frac{27}{2}$$

The equation is now solved.

$$x=-\sqrt{145}-\frac{27}{2} x=\sqrt{145}-\frac{27}{2}$$

Steps for Completing the Square

Consider \left(1+2x\right)\left(1-2x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$3x+1-\left(2x\right)^{2}=61x+50\left(x+3\right)$$

Expand \left(2x\right)^{2}.

$$3x+1-2^{2}x^{2}=61x+50\left(x+3\right)$$

Calculate 2 to the power of 2 and get 4.

$$3x+1-4x^{2}=61x+50\left(x+3\right)$$

Use the distributive property to multiply 50 by x+3.

$$3x+1-4x^{2}=61x+50x+150$$

Combine 61x and 50x to get 111x.

$$3x+1-4x^{2}=111x+150$$

Subtract 111x from both sides.

$$3x+1-4x^{2}-111x=150$$

Combine 3x and -111x to get -108x.

$$-108x+1-4x^{2}=150$$

Subtract 1 from both sides.

$$-108x-4x^{2}=150-1$$

Subtract 1 from 150 to get 149.

$$-108x-4x^{2}=149$$

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.

$$-4x^{2}-108x=149$$

Divide both sides by -4.

$$\frac{-4x^{2}-108x}{-4}=\frac{149}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{108}{-4}\right)x=\frac{149}{-4}$$

Divide -108 by -4.

$$x^{2}+27x=\frac{149}{-4}$$

Divide 149 by -4.

$$x^{2}+27x=-\frac{149}{4}$$

Divide 27, the coefficient of the x term, by 2 to get \frac{27}{2}. Then add the square of \frac{27}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+27x+\left(\frac{27}{2}\right)^{2}=-\frac{149}{4}+\left(\frac{27}{2}\right)^{2}$$

Square \frac{27}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+27x+\frac{729}{4}=\frac{-149+729}{4}$$

Add -\frac{149}{4} to \frac{729}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+27x+\frac{729}{4}=145$$

Factor x^{2}+27x+\frac{729}{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}.

$$\left(x+\frac{27}{2}\right)^{2}=145$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{27}{2}\right)^{2}}=\sqrt{145}$$

Simplify.

$$x+\frac{27}{2}=\sqrt{145} x+\frac{27}{2}=-\sqrt{145}$$

Subtract \frac{27}{2} from both sides of the equation.

$$x=\sqrt{145}-\frac{27}{2} x=-\sqrt{145}-\frac{27}{2}$$