Solve (5x-4)*(5x+4)/4=(6x-2)^2/4-18/4

$$\frac { ( 5 x - 4 ) \cdot ( 5 x + 4 ) } { 4 } = \frac { ( 6 x - 2 ) ^ { 2 } } { 4 } - \frac { 18 } { 4 }$$

Solve for x

$$x = \frac{\sqrt{122} + 12}{11} \approx 2.09503282$$

$$x=\frac{12-\sqrt{122}}{11}\approx 0.086785362$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$\left(5x-4\right)\left(5x+4\right)=\left(6x-2\right)^{2}-18$$

Consider \left(5x-4\right)\left(5x+4\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 4.

$$\left(5x\right)^{2}-16=\left(6x-2\right)^{2}-18$$

Expand \left(5x\right)^{2}.

$$5^{2}x^{2}-16=\left(6x-2\right)^{2}-18$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}-16=\left(6x-2\right)^{2}-18$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-2\right)^{2}.

$$25x^{2}-16=36x^{2}-24x+4-18$$

Subtract 18 from 4 to get -14.

$$25x^{2}-16=36x^{2}-24x-14$$

Subtract 36x^{2} from both sides.

$$25x^{2}-16-36x^{2}=-24x-14$$

Combine 25x^{2} and -36x^{2} to get -11x^{2}.

$$-11x^{2}-16=-24x-14$$

Add 24x to both sides.

$$-11x^{2}-16+24x=-14$$

Add 14 to both sides.

$$-11x^{2}-16+24x+14=0$$

Add -16 and 14 to get -2.

$$-11x^{2}-2+24x=0$$

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.

$$-11x^{2}+24x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -11 for a, 24 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\left(-11\right)\left(-2\right)}}{2\left(-11\right)}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\left(-11\right)\left(-2\right)}}{2\left(-11\right)}$$

Multiply -4 times -11.

$$x=\frac{-24±\sqrt{576+44\left(-2\right)}}{2\left(-11\right)}$$

Multiply 44 times -2.

$$x=\frac{-24±\sqrt{576-88}}{2\left(-11\right)}$$

Add 576 to -88.

$$x=\frac{-24±\sqrt{488}}{2\left(-11\right)}$$

Take the square root of 488.

$$x=\frac{-24±2\sqrt{122}}{2\left(-11\right)}$$

Multiply 2 times -11.

$$x=\frac{-24±2\sqrt{122}}{-22}$$

Now solve the equation x=\frac{-24±2\sqrt{122}}{-22} when ± is plus. Add -24 to 2\sqrt{122}.

$$x=\frac{2\sqrt{122}-24}{-22}$$

Divide -24+2\sqrt{122} by -22.

$$x=\frac{12-\sqrt{122}}{11}$$

Now solve the equation x=\frac{-24±2\sqrt{122}}{-22} when ± is minus. Subtract 2\sqrt{122} from -24.

$$x=\frac{-2\sqrt{122}-24}{-22}$$

Divide -24-2\sqrt{122} by -22.

$$x=\frac{\sqrt{122}+12}{11}$$

The equation is now solved.

$$x=\frac{12-\sqrt{122}}{11} x=\frac{\sqrt{122}+12}{11}$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$\left(5x-4\right)\left(5x+4\right)=\left(6x-2\right)^{2}-18$$

Consider \left(5x-4\right)\left(5x+4\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 4.

$$\left(5x\right)^{2}-16=\left(6x-2\right)^{2}-18$$

Expand \left(5x\right)^{2}.

$$5^{2}x^{2}-16=\left(6x-2\right)^{2}-18$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}-16=\left(6x-2\right)^{2}-18$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-2\right)^{2}.

$$25x^{2}-16=36x^{2}-24x+4-18$$

Subtract 18 from 4 to get -14.

$$25x^{2}-16=36x^{2}-24x-14$$

Subtract 36x^{2} from both sides.

$$25x^{2}-16-36x^{2}=-24x-14$$

Combine 25x^{2} and -36x^{2} to get -11x^{2}.

$$-11x^{2}-16=-24x-14$$

Add 24x to both sides.

$$-11x^{2}-16+24x=-14$$

Add 16 to both sides.

$$-11x^{2}+24x=-14+16$$

Add -14 and 16 to get 2.

$$-11x^{2}+24x=2$$

Divide both sides by -11.

$$\frac{-11x^{2}+24x}{-11}=\frac{2}{-11}$$

Dividing by -11 undoes the multiplication by -11.

$$x^{2}+\frac{24}{-11}x=\frac{2}{-11}$$

Divide 24 by -11.

$$x^{2}-\frac{24}{11}x=\frac{2}{-11}$$

Divide 2 by -11.

$$x^{2}-\frac{24}{11}x=-\frac{2}{11}$$

Divide -\frac{24}{11}, the coefficient of the x term, by 2 to get -\frac{12}{11}. Then add the square of -\frac{12}{11} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{24}{11}x+\left(-\frac{12}{11}\right)^{2}=-\frac{2}{11}+\left(-\frac{12}{11}\right)^{2}$$

Square -\frac{12}{11} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{24}{11}x+\frac{144}{121}=-\frac{2}{11}+\frac{144}{121}$$

Add -\frac{2}{11} to \frac{144}{121} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{24}{11}x+\frac{144}{121}=\frac{122}{121}$$

Factor x^{2}-\frac{24}{11}x+\frac{144}{121}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{12}{11}\right)^{2}=\frac{122}{121}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{12}{11}\right)^{2}}=\sqrt{\frac{122}{121}}$$

Simplify.

$$x-\frac{12}{11}=\frac{\sqrt{122}}{11} x-\frac{12}{11}=-\frac{\sqrt{122}}{11}$$

Add \frac{12}{11} to both sides of the equation.

$$x=\frac{\sqrt{122}+12}{11} x=\frac{12-\sqrt{122}}{11}$$