Solve 1(2x-5)^2-0.9(x+4)^2=0

$$1 ( 2 x - 5 ) ^ { 2 } - 0.9 ( x + 4 ) ^ { 2 } = 0$$

Solve for x

$$x = \frac{39 \sqrt{10} + 136}{31} \approx 8.365446089$$

$$x=\frac{136-39\sqrt{10}}{31}\approx 0.40874746$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$1\left(4x^{2}-20x+25\right)-0.9\left(x+4\right)^{2}=0$$

Use the distributive property to multiply 1 by 4x^{2}-20x+25.

$$4x^{2}-20x+25-0.9\left(x+4\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$4x^{2}-20x+25-0.9\left(x^{2}+8x+16\right)=0$$

Use the distributive property to multiply -0.9 by x^{2}+8x+16.

$$4x^{2}-20x+25-0.9x^{2}-7.2x-14.4=0$$

Combine 4x^{2} and -0.9x^{2} to get 3.1x^{2}.

$$3.1x^{2}-20x+25-7.2x-14.4=0$$

Combine -20x and -7.2x to get -27.2x.

$$3.1x^{2}-27.2x+25-14.4=0$$

Subtract 14.4 from 25 to get 10.6.

$$3.1x^{2}-27.2x+10.6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3.1 for a, -27.2 for b, and 10.6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-27.2\right)±\sqrt{\left(-27.2\right)^{2}-4\times 3.1\times 10.6}}{2\times 3.1}$$

Square -27.2 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-27.2\right)±\sqrt{739.84-4\times 3.1\times 10.6}}{2\times 3.1}$$

Multiply -4 times 3.1.

$$x=\frac{-\left(-27.2\right)±\sqrt{739.84-12.4\times 10.6}}{2\times 3.1}$$

Multiply -12.4 times 10.6 by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-27.2\right)±\sqrt{\frac{18496-3286}{25}}}{2\times 3.1}$$

Add 739.84 to -131.44 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-27.2\right)±\sqrt{608.4}}{2\times 3.1}$$

Take the square root of 608.4.

$$x=\frac{-\left(-27.2\right)±\frac{39\sqrt{10}}{5}}{2\times 3.1}$$

The opposite of -27.2 is 27.2.

$$x=\frac{27.2±\frac{39\sqrt{10}}{5}}{2\times 3.1}$$

Multiply 2 times 3.1.

$$x=\frac{27.2±\frac{39\sqrt{10}}{5}}{6.2}$$

Now solve the equation x=\frac{27.2±\frac{39\sqrt{10}}{5}}{6.2} when ± is plus. Add 27.2 to \frac{39\sqrt{10}}{5}.

$$x=\frac{39\sqrt{10}+136}{5\times 6.2}$$

Divide \frac{136+39\sqrt{10}}{5} by 6.2 by multiplying \frac{136+39\sqrt{10}}{5} by the reciprocal of 6.2.

$$x=\frac{39\sqrt{10}+136}{31}$$

Now solve the equation x=\frac{27.2±\frac{39\sqrt{10}}{5}}{6.2} when ± is minus. Subtract \frac{39\sqrt{10}}{5} from 27.2.

$$x=\frac{136-39\sqrt{10}}{5\times 6.2}$$

Divide \frac{136-39\sqrt{10}}{5} by 6.2 by multiplying \frac{136-39\sqrt{10}}{5} by the reciprocal of 6.2.

$$x=\frac{136-39\sqrt{10}}{31}$$

The equation is now solved.

$$x=\frac{39\sqrt{10}+136}{31} x=\frac{136-39\sqrt{10}}{31}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$1\left(4x^{2}-20x+25\right)-0.9\left(x+4\right)^{2}=0$$

Use the distributive property to multiply 1 by 4x^{2}-20x+25.

$$4x^{2}-20x+25-0.9\left(x+4\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$4x^{2}-20x+25-0.9\left(x^{2}+8x+16\right)=0$$

Use the distributive property to multiply -0.9 by x^{2}+8x+16.

$$4x^{2}-20x+25-0.9x^{2}-7.2x-14.4=0$$

Combine 4x^{2} and -0.9x^{2} to get 3.1x^{2}.

$$3.1x^{2}-20x+25-7.2x-14.4=0$$

Combine -20x and -7.2x to get -27.2x.

$$3.1x^{2}-27.2x+25-14.4=0$$

Subtract 14.4 from 25 to get 10.6.

$$3.1x^{2}-27.2x+10.6=0$$

Subtract 10.6 from both sides. Anything subtracted from zero gives its negation.

$$3.1x^{2}-27.2x=-10.6$$

Divide both sides of the equation by 3.1, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{3.1x^{2}-27.2x}{3.1}=-\frac{10.6}{3.1}$$

Dividing by 3.1 undoes the multiplication by 3.1.

$$x^{2}+\left(-\frac{27.2}{3.1}\right)x=-\frac{10.6}{3.1}$$

Divide -27.2 by 3.1 by multiplying -27.2 by the reciprocal of 3.1.

$$x^{2}-\frac{272}{31}x=-\frac{10.6}{3.1}$$

Divide -10.6 by 3.1 by multiplying -10.6 by the reciprocal of 3.1.

$$x^{2}-\frac{272}{31}x=-\frac{106}{31}$$

Divide -\frac{272}{31}, the coefficient of the x term, by 2 to get -\frac{136}{31}. Then add the square of -\frac{136}{31} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{272}{31}x+\left(-\frac{136}{31}\right)^{2}=-\frac{106}{31}+\left(-\frac{136}{31}\right)^{2}$$

Square -\frac{136}{31} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{272}{31}x+\frac{18496}{961}=-\frac{106}{31}+\frac{18496}{961}$$

Add -\frac{106}{31} to \frac{18496}{961} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{272}{31}x+\frac{18496}{961}=\frac{15210}{961}$$

Factor x^{2}-\frac{272}{31}x+\frac{18496}{961}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{136}{31}\right)^{2}=\frac{15210}{961}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{136}{31}\right)^{2}}=\sqrt{\frac{15210}{961}}$$

Simplify.

$$x-\frac{136}{31}=\frac{39\sqrt{10}}{31} x-\frac{136}{31}=-\frac{39\sqrt{10}}{31}$$

Add \frac{136}{31} to both sides of the equation.

$$x=\frac{39\sqrt{10}+136}{31} x=\frac{136-39\sqrt{10}}{31}$$