Solve 25-(2k+3)*5+k^2+3k+2=0

$$25 - ( 2 k + 3 ) \times 5 + k ^ { 2 } + 3 k + 2 = 0$$

Solve for k

$$k=3$$

$$k=4$$

Steps Using Factoring

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

$$25-\left(10k+15\right)+k^{2}+3k+2=0$$

To find the opposite of 10k+15, find the opposite of each term.

$$25-10k-15+k^{2}+3k+2=0$$

Subtract 15 from 25 to get 10.

$$10-10k+k^{2}+3k+2=0$$

Combine -10k and 3k to get -7k.

$$10-7k+k^{2}+2=0$$

Add 10 and 2 to get 12.

$$12-7k+k^{2}=0$$

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

$$k^{2}-7k+12=0$$

To solve the equation, factor k^{2}-7k+12 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+b\right). To find a and b, set up a system to be solved.

$$a+b=-7 ab=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite factored expression \left(k+a\right)\left(k+b\right) using the obtained values.

$$\left(k-4\right)\left(k-3\right)$$

To find equation solutions, solve k-4=0 and k-3=0.

$$k=4 k=3$$

Steps Using Factoring By Grouping

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

$$25-\left(10k+15\right)+k^{2}+3k+2=0$$

To find the opposite of 10k+15, find the opposite of each term.

$$25-10k-15+k^{2}+3k+2=0$$

Subtract 15 from 25 to get 10.

$$10-10k+k^{2}+3k+2=0$$

Combine -10k and 3k to get -7k.

$$10-7k+k^{2}+2=0$$

Add 10 and 2 to get 12.

$$12-7k+k^{2}=0$$

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

$$k^{2}-7k+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk+12. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite k^{2}-7k+12 as \left(k^{2}-4k\right)+\left(-3k+12\right).

$$\left(k^{2}-4k\right)+\left(-3k+12\right)$$

Factor out k in the first and -3 in the second group.

$$k\left(k-4\right)-3\left(k-4\right)$$

Factor out common term k-4 by using distributive property.

$$\left(k-4\right)\left(k-3\right)$$

To find equation solutions, solve k-4=0 and k-3=0.

$$k=4 k=3$$

Steps Using the Quadratic Formula

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

$$25-\left(10k+15\right)+k^{2}+3k+2=0$$

To find the opposite of 10k+15, find the opposite of each term.

$$25-10k-15+k^{2}+3k+2=0$$

Subtract 15 from 25 to get 10.

$$10-10k+k^{2}+3k+2=0$$

Combine -10k and 3k to get -7k.

$$10-7k+k^{2}+2=0$$

Add 10 and 2 to get 12.

$$12-7k+k^{2}=0$$

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.

$$k^{2}-7k+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12}}{2}$$

Square -7.

$$k=\frac{-\left(-7\right)±\sqrt{49-4\times 12}}{2}$$

Multiply -4 times 12.

$$k=\frac{-\left(-7\right)±\sqrt{49-48}}{2}$$

Add 49 to -48.

$$k=\frac{-\left(-7\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$k=\frac{-\left(-7\right)±1}{2}$$

The opposite of -7 is 7.

$$k=\frac{7±1}{2}$$

Now solve the equation k=\frac{7±1}{2} when ± is plus. Add 7 to 1.

$$k=\frac{8}{2}$$

Divide 8 by 2.

$$k=4$$

Now solve the equation k=\frac{7±1}{2} when ± is minus. Subtract 1 from 7.

$$k=\frac{6}{2}$$

Divide 6 by 2.

$$k=3$$

The equation is now solved.

$$k=4 k=3$$

Steps for Completing the Square

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

$$25-\left(10k+15\right)+k^{2}+3k+2=0$$

To find the opposite of 10k+15, find the opposite of each term.

$$25-10k-15+k^{2}+3k+2=0$$

Subtract 15 from 25 to get 10.

$$10-10k+k^{2}+3k+2=0$$

Combine -10k and 3k to get -7k.

$$10-7k+k^{2}+2=0$$

Add 10 and 2 to get 12.

$$12-7k+k^{2}=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$-7k+k^{2}=-12$$

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.

$$k^{2}-7k=-12$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-7k+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-7k+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$k^{2}-7k+\frac{49}{4}=\frac{1}{4}$$

Factor k^{2}-7k+\frac{49}{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}.

$$\left(k-\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$k-\frac{7}{2}=\frac{1}{2} k-\frac{7}{2}=-\frac{1}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$k=4 k=3$$