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a+b=3 ab=5\left(-2\right)=-10
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-2. To find a and b, set up a system to be solved.
-1,10 -2,5
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 -10.
-1+10=9 -2+5=3
Calculate the sum for each pair.
a=-2 b=5
The solution is the pair that gives sum 3.
\left(5x^{2}-2x\right)+\left(5x-2\right)
Rewrite 5x^{2}+3x-2 as \left(5x^{2}-2x\right)+\left(5x-2\right).
x\left(5x-2\right)+5x-2
Factor out x in 5x^{2}-2x.
\left(5x-2\right)\left(x+1\right)
Factor out common term 5x-2 by using distributive property.
x=\frac{2}{5} x=-1
To find equation solutions, solve 5x-2=0 and x+1=0.
5x^{2}+3x-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.
x=\frac{-3±\sqrt{3^{2}-4\times 5\left(-2\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±\sqrt{9-4\times 5\left(-2\right)}}{2\times 5}
Square 3.
x=\frac{-3±\sqrt{9-20\left(-2\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-3±\sqrt{9+40}}{2\times 5}
Multiply -20 times -2.
x=\frac{-3±\sqrt{49}}{2\times 5}
Add 9 to 40.
x=\frac{-3±7}{2\times 5}
Take the square root of 49.
x=\frac{-3±7}{10}
Multiply 2 times 5.
x=\frac{4}{10}
Now solve the equation x=\frac{-3±7}{10} when ± is plus. Add -3 to 7.
x=\frac{2}{5}
Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.
x=-\frac{10}{10}
Now solve the equation x=\frac{-3±7}{10} when ± is minus. Subtract 7 from -3.
x=-1
Divide -10 by 10.
x=\frac{2}{5} x=-1
The equation is now solved.
5x^{2}+3x-2=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}+3x-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
5x^{2}+3x=-\left(-2\right)
Subtracting -2 from itself leaves 0.
5x^{2}+3x=2
Subtract -2 from 0.
\frac{5x^{2}+3x}{5}=\frac{2}{5}
Divide both sides by 5.
x^{2}+\frac{3}{5}x=\frac{2}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}+\frac{3}{5}x+\left(\frac{3}{10}\right)^{2}=\frac{2}{5}+\left(\frac{3}{10}\right)^{2}
Divide \frac{3}{5}, the coefficient of the x term, by 2 to get \frac{3}{10}. Then add the square of \frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{2}{5}+\frac{9}{100}
Square \frac{3}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{3}{5}x+\frac{9}{100}=\frac{49}{100}
Add \frac{2}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{3}{10}\right)^{2}=\frac{49}{100}
Factor x^{2}+\frac{3}{5}x+\frac{9}{100}. 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{3}{10}\right)^{2}}=\sqrt{\frac{49}{100}}
Take the square root of both sides of the equation.
x+\frac{3}{10}=\frac{7}{10} x+\frac{3}{10}=-\frac{7}{10}
Simplify.
x=\frac{2}{5} x=-1
Subtract \frac{3}{10} from both sides of the equation.
x ^ 2 +\frac{3}{5}x -\frac{2}{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{3}{5} rs = -\frac{2}{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{3}{10} - u s = -\frac{3}{10} + u
Two numbers r and s sum up to -\frac{3}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{5} = -\frac{3}{10}. 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{3}{10} - u) (-\frac{3}{10} + u) = -\frac{2}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{5}
\frac{9}{100} - u^2 = -\frac{2}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{2}{5}-\frac{9}{100} = -\frac{49}{100}
Simplify the expression by subtracting \frac{9}{100} on both sides
u^2 = \frac{49}{100} u = \pm\sqrt{\frac{49}{100}} = \pm \frac{7}{10}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{3}{10} - \frac{7}{10} = -1 s = -\frac{3}{10} + \frac{7}{10} = 0.400
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.