Solve for x

x=1

x=\frac{3}{5}=0.6

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a+b=-8 ab=5\times 3=15

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

-1,-15 -3,-5

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 15.

-1-15=-16 -3-5=-8

Calculate the sum for each pair.

a=-5 b=-3

The solution is the pair that gives sum -8.

\left(5x^{2}-5x\right)+\left(-3x+3\right)

Rewrite 5x^{2}-8x+3 as \left(5x^{2}-5x\right)+\left(-3x+3\right).

5x\left(x-1\right)-3\left(x-1\right)

Factor out 5x in the first and -3 in the second group.

\left(x-1\right)\left(5x-3\right)

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

x=1 x=\frac{3}{5}

To find equation solutions, solve x-1=0 and 5x-3=0.

5x^{2}-8x+3=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.

x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 5\times 3}}{2\times 5}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -8 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-8\right)±\sqrt{64-4\times 5\times 3}}{2\times 5}

Square -8.

x=\frac{-\left(-8\right)±\sqrt{64-20\times 3}}{2\times 5}

Multiply -4 times 5.

x=\frac{-\left(-8\right)±\sqrt{64-60}}{2\times 5}

Multiply -20 times 3.

x=\frac{-\left(-8\right)±\sqrt{4}}{2\times 5}

Add 64 to -60.

x=\frac{-\left(-8\right)±2}{2\times 5}

Take the square root of 4.

x=\frac{8±2}{2\times 5}

The opposite of -8 is 8.

x=\frac{8±2}{10}

Multiply 2 times 5.

x=\frac{10}{10}

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

x=1

Divide 10 by 10.

x=\frac{6}{10}

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

x=\frac{3}{5}

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

x=1 x=\frac{3}{5}

The equation is now solved.

5x^{2}-8x+3=0

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.

5x^{2}-8x+3-3=-3

Subtract 3 from both sides of the equation.

5x^{2}-8x=-3

Subtracting 3 from itself leaves 0.

\frac{5x^{2}-8x}{5}=\frac{-3}{5}

Divide both sides by 5.

x^{2}+\frac{-8}{5}x=\frac{-3}{5}

Dividing by 5 undoes the multiplication by 5.

x^{2}-\frac{8}{5}x=\frac{-3}{5}

Divide -8 by 5.

x^{2}-\frac{8}{5}x=-\frac{3}{5}

Divide -3 by 5.

x^{2}-\frac{8}{5}x+\left(-\frac{4}{5}\right)^{2}=-\frac{3}{5}+\left(-\frac{4}{5}\right)^{2}

Divide -\frac{8}{5}, the coefficient of the x term, by 2 to get -\frac{4}{5}. Then add the square of -\frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}-\frac{8}{5}x+\frac{16}{25}=-\frac{3}{5}+\frac{16}{25}

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

x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{1}{25}

Add -\frac{3}{5} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

\left(x-\frac{4}{5}\right)^{2}=\frac{1}{25}

Factor x^{2}-\frac{8}{5}x+\frac{16}{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}.

\sqrt{\left(x-\frac{4}{5}\right)^{2}}=\sqrt{\frac{1}{25}}

Take the square root of both sides of the equation.

x-\frac{4}{5}=\frac{1}{5} x-\frac{4}{5}=-\frac{1}{5}

Simplify.

x=1 x=\frac{3}{5}

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

x ^ 2 -\frac{8}{5}x +\frac{3}{5} = 0

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 5

r + s = \frac{8}{5} rs = \frac{3}{5}

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 = \frac{4}{5} - u s = \frac{4}{5} + u

Two numbers r and s sum up to \frac{8}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{8}{5} = \frac{4}{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>

(\frac{4}{5} - u) (\frac{4}{5} + u) = \frac{3}{5}

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{5}

\frac{16}{25} - u^2 = \frac{3}{5}

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

-u^2 = \frac{3}{5}-\frac{16}{25} = -\frac{1}{25}

Simplify the expression by subtracting \frac{16}{25} on both sides

u^2 = \frac{1}{25} u = \pm\sqrt{\frac{1}{25}} = \pm \frac{1}{5}

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

r =\frac{4}{5} - \frac{1}{5} = 0.600 s = \frac{4}{5} + \frac{1}{5} = 1.000

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

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