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a+b=23 ab=15\left(-28\right)=-420
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15x^{2}+ax+bx-28. To find a and b, set up a system to be solved.
-1,420 -2,210 -3,140 -4,105 -5,84 -6,70 -7,60 -10,42 -12,35 -14,30 -15,28 -20,21
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 -420.
-1+420=419 -2+210=208 -3+140=137 -4+105=101 -5+84=79 -6+70=64 -7+60=53 -10+42=32 -12+35=23 -14+30=16 -15+28=13 -20+21=1
Calculate the sum for each pair.
a=-12 b=35
The solution is the pair that gives sum 23.
\left(15x^{2}-12x\right)+\left(35x-28\right)
Rewrite 15x^{2}+23x-28 as \left(15x^{2}-12x\right)+\left(35x-28\right).
3x\left(5x-4\right)+7\left(5x-4\right)
Factor out 3x in the first and 7 in the second group.
\left(5x-4\right)\left(3x+7\right)
Factor out common term 5x-4 by using distributive property.
x=\frac{4}{5} x=-\frac{7}{3}
To find equation solutions, solve 5x-4=0 and 3x+7=0.
15x^{2}+23x-28=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{-23±\sqrt{23^{2}-4\times 15\left(-28\right)}}{2\times 15}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, 23 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-23±\sqrt{529-4\times 15\left(-28\right)}}{2\times 15}
Square 23.
x=\frac{-23±\sqrt{529-60\left(-28\right)}}{2\times 15}
Multiply -4 times 15.
x=\frac{-23±\sqrt{529+1680}}{2\times 15}
Multiply -60 times -28.
x=\frac{-23±\sqrt{2209}}{2\times 15}
Add 529 to 1680.
x=\frac{-23±47}{2\times 15}
Take the square root of 2209.
x=\frac{-23±47}{30}
Multiply 2 times 15.
x=\frac{24}{30}
Now solve the equation x=\frac{-23±47}{30} when ± is plus. Add -23 to 47.
x=\frac{4}{5}
Reduce the fraction \frac{24}{30} to lowest terms by extracting and canceling out 6.
x=-\frac{70}{30}
Now solve the equation x=\frac{-23±47}{30} when ± is minus. Subtract 47 from -23.
x=-\frac{7}{3}
Reduce the fraction \frac{-70}{30} to lowest terms by extracting and canceling out 10.
x=\frac{4}{5} x=-\frac{7}{3}
The equation is now solved.
15x^{2}+23x-28=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.
15x^{2}+23x-28-\left(-28\right)=-\left(-28\right)
Add 28 to both sides of the equation.
15x^{2}+23x=-\left(-28\right)
Subtracting -28 from itself leaves 0.
15x^{2}+23x=28
Subtract -28 from 0.
\frac{15x^{2}+23x}{15}=\frac{28}{15}
Divide both sides by 15.
x^{2}+\frac{23}{15}x=\frac{28}{15}
Dividing by 15 undoes the multiplication by 15.
x^{2}+\frac{23}{15}x+\left(\frac{23}{30}\right)^{2}=\frac{28}{15}+\left(\frac{23}{30}\right)^{2}
Divide \frac{23}{15}, the coefficient of the x term, by 2 to get \frac{23}{30}. Then add the square of \frac{23}{30} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{23}{15}x+\frac{529}{900}=\frac{28}{15}+\frac{529}{900}
Square \frac{23}{30} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{23}{15}x+\frac{529}{900}=\frac{2209}{900}
Add \frac{28}{15} to \frac{529}{900} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{23}{30}\right)^{2}=\frac{2209}{900}
Factor x^{2}+\frac{23}{15}x+\frac{529}{900}. 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{23}{30}\right)^{2}}=\sqrt{\frac{2209}{900}}
Take the square root of both sides of the equation.
x+\frac{23}{30}=\frac{47}{30} x+\frac{23}{30}=-\frac{47}{30}
Simplify.
x=\frac{4}{5} x=-\frac{7}{3}
Subtract \frac{23}{30} from both sides of the equation.
x ^ 2 +\frac{23}{15}x -\frac{28}{15} = 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 15
r + s = -\frac{23}{15} rs = -\frac{28}{15}
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{23}{30} - u s = -\frac{23}{30} + u
Two numbers r and s sum up to -\frac{23}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{23}{15} = -\frac{23}{30}. 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{23}{30} - u) (-\frac{23}{30} + u) = -\frac{28}{15}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{28}{15}
\frac{529}{900} - u^2 = -\frac{28}{15}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{28}{15}-\frac{529}{900} = -\frac{2209}{900}
Simplify the expression by subtracting \frac{529}{900} on both sides
u^2 = \frac{2209}{900} u = \pm\sqrt{\frac{2209}{900}} = \pm \frac{47}{30}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{23}{30} - \frac{47}{30} = -2.333 s = -\frac{23}{30} + \frac{47}{30} = 0.800
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.