Solve 3/2x-3+14/4x^2+4x-15+5=0

$$\frac { 3 } { 2 x - 3 } + \frac { 14 } { 4 x ^ { 2 } + 4 x - 15 } + 5 = 0$$

Solve for x

$$x = -\frac{23}{10} = -2\frac{3}{10} = -2.3$$

$$x=1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -\frac{5}{2},\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-3\right)\left(2x+5\right), the least common multiple of 2x-3,4x^{2}+4x-15.

$$\left(2x+5\right)\times 3+14+\left(2x-3\right)\left(2x+5\right)\times 5=0$$

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

$$6x+15+14+\left(2x-3\right)\left(2x+5\right)\times 5=0$$

Add 15 and 14 to get 29.

$$6x+29+\left(2x-3\right)\left(2x+5\right)\times 5=0$$

Use the distributive property to multiply 2x-3 by 2x+5 and combine like terms.

$$6x+29+\left(4x^{2}+4x-15\right)\times 5=0$$

Use the distributive property to multiply 4x^{2}+4x-15 by 5.

$$6x+29+20x^{2}+20x-75=0$$

Combine 6x and 20x to get 26x.

$$26x+29+20x^{2}-75=0$$

Subtract 75 from 29 to get -46.

$$26x-46+20x^{2}=0$$

Divide both sides by 2.

$$13x-23+10x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$10x^{2}+13x-23=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx-23. To find a and b, set up a system to be solved.

$$a+b=13 ab=10\left(-23\right)=-230$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -230.

$$-1,230 -2,115 -5,46 -10,23$$

Calculate the sum for each pair.

$$-1+230=229 -2+115=113 -5+46=41 -10+23=13$$

The solution is the pair that gives sum 13.

$$a=-10 b=23$$

Rewrite 10x^{2}+13x-23 as \left(10x^{2}-10x\right)+\left(23x-23\right).

$$\left(10x^{2}-10x\right)+\left(23x-23\right)$$

Factor out 10x in the first and 23 in the second group.

$$10x\left(x-1\right)+23\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(10x+23\right)$$

To find equation solutions, solve x-1=0 and 10x+23=0.

$$x=1 x=-\frac{23}{10}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{5}{2},\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-3\right)\left(2x+5\right), the least common multiple of 2x-3,4x^{2}+4x-15.

$$\left(2x+5\right)\times 3+14+\left(2x-3\right)\left(2x+5\right)\times 5=0$$

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

$$6x+15+14+\left(2x-3\right)\left(2x+5\right)\times 5=0$$

Add 15 and 14 to get 29.

$$6x+29+\left(2x-3\right)\left(2x+5\right)\times 5=0$$

Use the distributive property to multiply 2x-3 by 2x+5 and combine like terms.

$$6x+29+\left(4x^{2}+4x-15\right)\times 5=0$$

Use the distributive property to multiply 4x^{2}+4x-15 by 5.

$$6x+29+20x^{2}+20x-75=0$$

Combine 6x and 20x to get 26x.

$$26x+29+20x^{2}-75=0$$

Subtract 75 from 29 to get -46.

$$26x-46+20x^{2}=0$$

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.

$$20x^{2}+26x-46=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, 26 for b, and -46 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-26±\sqrt{26^{2}-4\times 20\left(-46\right)}}{2\times 20}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\times 20\left(-46\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-26±\sqrt{676-80\left(-46\right)}}{2\times 20}$$

Multiply -80 times -46.

$$x=\frac{-26±\sqrt{676+3680}}{2\times 20}$$

Add 676 to 3680.

$$x=\frac{-26±\sqrt{4356}}{2\times 20}$$

Take the square root of 4356.

$$x=\frac{-26±66}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-26±66}{40}$$

Now solve the equation x=\frac{-26±66}{40} when ± is plus. Add -26 to 66.

$$x=\frac{40}{40}$$

Divide 40 by 40.

$$x=1$$

Now solve the equation x=\frac{-26±66}{40} when ± is minus. Subtract 66 from -26.

$$x=-\frac{92}{40}$$

Reduce the fraction \frac{-92}{40} to lowest terms by extracting and canceling out 4.

$$x=-\frac{23}{10}$$

The equation is now solved.

$$x=1 x=-\frac{23}{10}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{5}{2},\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-3\right)\left(2x+5\right), the least common multiple of 2x-3,4x^{2}+4x-15.

$$\left(2x+5\right)\times 3+14+\left(2x-3\right)\left(2x+5\right)\times 5=0$$

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

$$6x+15+14+\left(2x-3\right)\left(2x+5\right)\times 5=0$$

Add 15 and 14 to get 29.

$$6x+29+\left(2x-3\right)\left(2x+5\right)\times 5=0$$

Use the distributive property to multiply 2x-3 by 2x+5 and combine like terms.

$$6x+29+\left(4x^{2}+4x-15\right)\times 5=0$$

Use the distributive property to multiply 4x^{2}+4x-15 by 5.

$$6x+29+20x^{2}+20x-75=0$$

Combine 6x and 20x to get 26x.

$$26x+29+20x^{2}-75=0$$

Subtract 75 from 29 to get -46.

$$26x-46+20x^{2}=0$$

Add 46 to both sides. Anything plus zero gives itself.

$$26x+20x^{2}=46$$

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.

$$20x^{2}+26x=46$$

Divide both sides by 20.

$$\frac{20x^{2}+26x}{20}=\frac{46}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\frac{26}{20}x=\frac{46}{20}$$

Reduce the fraction \frac{26}{20} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{13}{10}x=\frac{46}{20}$$

Reduce the fraction \frac{46}{20} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{13}{10}x=\frac{23}{10}$$

Divide \frac{13}{10}, the coefficient of the x term, by 2 to get \frac{13}{20}. Then add the square of \frac{13}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{10}x+\left(\frac{13}{20}\right)^{2}=\frac{23}{10}+\left(\frac{13}{20}\right)^{2}$$

Square \frac{13}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{10}x+\frac{169}{400}=\frac{23}{10}+\frac{169}{400}$$

Add \frac{23}{10} to \frac{169}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{13}{10}x+\frac{169}{400}=\frac{1089}{400}$$

Factor x^{2}+\frac{13}{10}x+\frac{169}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{13}{20}\right)^{2}=\frac{1089}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{20}\right)^{2}}=\sqrt{\frac{1089}{400}}$$

Simplify.

$$x+\frac{13}{20}=\frac{33}{20} x+\frac{13}{20}=-\frac{33}{20}$$

Subtract \frac{13}{20} from both sides of the equation.

$$x=1 x=-\frac{23}{10}$$