Solve 50j^2+60j+18=0

$$50 j ^ { 2 } + 60 j + 18 = 0$$

Solve for j

$$j=-\frac{3}{5}=-0.6$$

Steps Using Factoring By Grouping

Divide both sides by 2.

$$25j^{2}+30j+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25j^{2}+aj+bj+9. To find a and b, set up a system to be solved.

$$a+b=30 ab=25\times 9=225$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 225.

$$1,225 3,75 5,45 9,25 15,15$$

Calculate the sum for each pair.

$$1+225=226 3+75=78 5+45=50 9+25=34 15+15=30$$

The solution is the pair that gives sum 30.

$$a=15 b=15$$

Rewrite 25j^{2}+30j+9 as \left(25j^{2}+15j\right)+\left(15j+9\right).

$$\left(25j^{2}+15j\right)+\left(15j+9\right)$$

Factor out 5j in the first and 3 in the second group.

$$5j\left(5j+3\right)+3\left(5j+3\right)$$

Factor out common term 5j+3 by using distributive property.

$$\left(5j+3\right)\left(5j+3\right)$$

Rewrite as a binomial square.

$$\left(5j+3\right)^{2}$$

To find equation solution, solve 5j+3=0.

$$j=-\frac{3}{5}$$

Steps Using the Quadratic Formula

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.

$$50j^{2}+60j+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 50 for a, 60 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$j=\frac{-60±\sqrt{60^{2}-4\times 50\times 18}}{2\times 50}$$

Square 60.

$$j=\frac{-60±\sqrt{3600-4\times 50\times 18}}{2\times 50}$$

Multiply -4 times 50.

$$j=\frac{-60±\sqrt{3600-200\times 18}}{2\times 50}$$

Multiply -200 times 18.

$$j=\frac{-60±\sqrt{3600-3600}}{2\times 50}$$

Add 3600 to -3600.

$$j=\frac{-60±\sqrt{0}}{2\times 50}$$

Take the square root of 0.

$$j=-\frac{60}{2\times 50}$$

Multiply 2 times 50.

$$j=-\frac{60}{100}$$

Reduce the fraction \frac{-60}{100} to lowest terms by extracting and canceling out 20.

$$j=-\frac{3}{5}$$

Steps for Completing the Square

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.

$$50j^{2}+60j+18=0$$

Subtract 18 from both sides of the equation.

$$50j^{2}+60j+18-18=-18$$

Subtracting 18 from itself leaves 0.

$$50j^{2}+60j=-18$$

Divide both sides by 50.

$$\frac{50j^{2}+60j}{50}=-\frac{18}{50}$$

Dividing by 50 undoes the multiplication by 50.

$$j^{2}+\frac{60}{50}j=-\frac{18}{50}$$

Reduce the fraction \frac{60}{50} to lowest terms by extracting and canceling out 10.

$$j^{2}+\frac{6}{5}j=-\frac{18}{50}$$

Reduce the fraction \frac{-18}{50} to lowest terms by extracting and canceling out 2.

$$j^{2}+\frac{6}{5}j=-\frac{9}{25}$$

Divide \frac{6}{5}, the coefficient of the x term, by 2 to get \frac{3}{5}. Then add the square of \frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$j^{2}+\frac{6}{5}j+\left(\frac{3}{5}\right)^{2}=-\frac{9}{25}+\left(\frac{3}{5}\right)^{2}$$

Square \frac{3}{5} by squaring both the numerator and the denominator of the fraction.

$$j^{2}+\frac{6}{5}j+\frac{9}{25}=\frac{-9+9}{25}$$

Add -\frac{9}{25} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$j^{2}+\frac{6}{5}j+\frac{9}{25}=0$$

Factor j^{2}+\frac{6}{5}j+\frac{9}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(j+\frac{3}{5}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(j+\frac{3}{5}\right)^{2}}=\sqrt{0}$$

Simplify.

$$j+\frac{3}{5}=0 j+\frac{3}{5}=0$$

Subtract \frac{3}{5} from both sides of the equation.

$$j=-\frac{3}{5} j=-\frac{3}{5}$$

The equation is now solved. Solutions are the same.

$$j=-\frac{3}{5}$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 50

$$x ^ 2 +\frac{6}{5}x +\frac{9}{25} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{6}{5} rs = \frac{9}{25}$$

Two numbers r and s sum up to -\frac{6}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{6}{5} = -\frac{3}{5}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{3}{5} - u s = -\frac{3}{5} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{25}

$$(-\frac{3}{5} - u) (-\frac{3}{5} + u) = \frac{9}{25}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{9}{25} - u^2 = \frac{9}{25}$$

Simplify the expression by subtracting \frac{9}{25} on both sides

$$-u^2 = \frac{9}{25}-\frac{9}{25} = 0$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 0 u = 0$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r = s = -\frac{3}{5} = -0.600$$