Solve 3left(2x-4right)-left(2x-5right)*left(2x+5right)+26=1-4left(x+5right)

$$3 \left( 2x-4 \right) - \left( 2x-5 \right) \times \left( 2x+5 \right) +26 = 1-4 \left( x+5 \right)$$

Solve for x

$$x = \frac{\sqrt{257} + 5}{4} \approx 5.257804885$$

$$x=\frac{5-\sqrt{257}}{4}\approx -2.757804885$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by 2x-4.

$$6x-12-\left(2x-5\right)\left(2x+5\right)+26=1-4\left(x+5\right)$$

Consider \left(2x-5\right)\left(2x+5\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 5.

$$6x-12-\left(\left(2x\right)^{2}-25\right)+26=1-4\left(x+5\right)$$

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

$$6x-12-\left(2^{2}x^{2}-25\right)+26=1-4\left(x+5\right)$$

Calculate 2 to the power of 2 and get 4.

$$6x-12-\left(4x^{2}-25\right)+26=1-4\left(x+5\right)$$

To find the opposite of 4x^{2}-25, find the opposite of each term.

$$6x-12-4x^{2}+25+26=1-4\left(x+5\right)$$

Add -12 and 25 to get 13.

$$6x+13-4x^{2}+26=1-4\left(x+5\right)$$

Add 13 and 26 to get 39.

$$6x+39-4x^{2}=1-4\left(x+5\right)$$

Use the distributive property to multiply -4 by x+5.

$$6x+39-4x^{2}=1-4x-20$$

Subtract 20 from 1 to get -19.

$$6x+39-4x^{2}=-19-4x$$

Subtract -19 from both sides.

$$6x+39-4x^{2}-\left(-19\right)=-4x$$

The opposite of -19 is 19.

$$6x+39-4x^{2}+19=-4x$$

Add 4x to both sides.

$$6x+39-4x^{2}+19+4x=0$$

Add 39 and 19 to get 58.

$$6x+58-4x^{2}+4x=0$$

Combine 6x and 4x to get 10x.

$$10x+58-4x^{2}=0$$

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.

$$-4x^{2}+10x+58=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 10 for b, and 58 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-4\right)\times 58}}{2\left(-4\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-4\right)\times 58}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-10±\sqrt{100+16\times 58}}{2\left(-4\right)}$$

Multiply 16 times 58.

$$x=\frac{-10±\sqrt{100+928}}{2\left(-4\right)}$$

Add 100 to 928.

$$x=\frac{-10±\sqrt{1028}}{2\left(-4\right)}$$

Take the square root of 1028.

$$x=\frac{-10±2\sqrt{257}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-10±2\sqrt{257}}{-8}$$

Now solve the equation x=\frac{-10±2\sqrt{257}}{-8} when ± is plus. Add -10 to 2\sqrt{257}.

$$x=\frac{2\sqrt{257}-10}{-8}$$

Divide -10+2\sqrt{257} by -8.

$$x=\frac{5-\sqrt{257}}{4}$$

Now solve the equation x=\frac{-10±2\sqrt{257}}{-8} when ± is minus. Subtract 2\sqrt{257} from -10.

$$x=\frac{-2\sqrt{257}-10}{-8}$$

Divide -10-2\sqrt{257} by -8.

$$x=\frac{\sqrt{257}+5}{4}$$

The equation is now solved.

$$x=\frac{5-\sqrt{257}}{4} x=\frac{\sqrt{257}+5}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 3 by 2x-4.

$$6x-12-\left(2x-5\right)\left(2x+5\right)+26=1-4\left(x+5\right)$$

Consider \left(2x-5\right)\left(2x+5\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 5.

$$6x-12-\left(\left(2x\right)^{2}-25\right)+26=1-4\left(x+5\right)$$

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

$$6x-12-\left(2^{2}x^{2}-25\right)+26=1-4\left(x+5\right)$$

Calculate 2 to the power of 2 and get 4.

$$6x-12-\left(4x^{2}-25\right)+26=1-4\left(x+5\right)$$

To find the opposite of 4x^{2}-25, find the opposite of each term.

$$6x-12-4x^{2}+25+26=1-4\left(x+5\right)$$

Add -12 and 25 to get 13.

$$6x+13-4x^{2}+26=1-4\left(x+5\right)$$

Add 13 and 26 to get 39.

$$6x+39-4x^{2}=1-4\left(x+5\right)$$

Use the distributive property to multiply -4 by x+5.

$$6x+39-4x^{2}=1-4x-20$$

Subtract 20 from 1 to get -19.

$$6x+39-4x^{2}=-19-4x$$

Add 4x to both sides.

$$6x+39-4x^{2}+4x=-19$$

Combine 6x and 4x to get 10x.

$$10x+39-4x^{2}=-19$$

Subtract 39 from both sides.

$$10x-4x^{2}=-19-39$$

Subtract 39 from -19 to get -58.

$$10x-4x^{2}=-58$$

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.

$$-4x^{2}+10x=-58$$

Divide both sides by -4.

$$\frac{-4x^{2}+10x}{-4}=-\frac{58}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{10}{-4}x=-\frac{58}{-4}$$

Reduce the fraction \frac{10}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{5}{2}x=-\frac{58}{-4}$$

Reduce the fraction \frac{-58}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{5}{2}x=\frac{29}{2}$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=\frac{29}{2}+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{29}{2}+\frac{25}{16}$$

Add \frac{29}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{257}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{257}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{257}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{\sqrt{257}}{4} x-\frac{5}{4}=-\frac{\sqrt{257}}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{\sqrt{257}+5}{4} x=\frac{5-\sqrt{257}}{4}$$