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a+b=17 ab=7\times 6=42
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx+6. To find a and b, set up a system to be solved.
1,42 2,21 3,14 6,7
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 42.
1+42=43 2+21=23 3+14=17 6+7=13
Calculate the sum for each pair.
a=3 b=14
The solution is the pair that gives sum 17.
\left(7x^{2}+3x\right)+\left(14x+6\right)
Rewrite 7x^{2}+17x+6 as \left(7x^{2}+3x\right)+\left(14x+6\right).
x\left(7x+3\right)+2\left(7x+3\right)
Factor out x in the first and 2 in the second group.
\left(7x+3\right)\left(x+2\right)
Factor out common term 7x+3 by using distributive property.
x=-\frac{3}{7} x=-2
To find equation solutions, solve 7x+3=0 and x+2=0.
7x^{2}+17x+6=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{-17±\sqrt{17^{2}-4\times 7\times 6}}{2\times 7}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 17 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-17±\sqrt{289-4\times 7\times 6}}{2\times 7}
Square 17.
x=\frac{-17±\sqrt{289-28\times 6}}{2\times 7}
Multiply -4 times 7.
x=\frac{-17±\sqrt{289-168}}{2\times 7}
Multiply -28 times 6.
x=\frac{-17±\sqrt{121}}{2\times 7}
Add 289 to -168.
x=\frac{-17±11}{2\times 7}
Take the square root of 121.
x=\frac{-17±11}{14}
Multiply 2 times 7.
x=-\frac{6}{14}
Now solve the equation x=\frac{-17±11}{14} when ± is plus. Add -17 to 11.
x=-\frac{3}{7}
Reduce the fraction \frac{-6}{14} to lowest terms by extracting and canceling out 2.
x=-\frac{28}{14}
Now solve the equation x=\frac{-17±11}{14} when ± is minus. Subtract 11 from -17.
x=-2
Divide -28 by 14.
x=-\frac{3}{7} x=-2
The equation is now solved.
7x^{2}+17x+6=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.
7x^{2}+17x+6-6=-6
Subtract 6 from both sides of the equation.
7x^{2}+17x=-6
Subtracting 6 from itself leaves 0.
\frac{7x^{2}+17x}{7}=-\frac{6}{7}
Divide both sides by 7.
x^{2}+\frac{17}{7}x=-\frac{6}{7}
Dividing by 7 undoes the multiplication by 7.
x^{2}+\frac{17}{7}x+\left(\frac{17}{14}\right)^{2}=-\frac{6}{7}+\left(\frac{17}{14}\right)^{2}
Divide \frac{17}{7}, the coefficient of the x term, by 2 to get \frac{17}{14}. Then add the square of \frac{17}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{17}{7}x+\frac{289}{196}=-\frac{6}{7}+\frac{289}{196}
Square \frac{17}{14} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{17}{7}x+\frac{289}{196}=\frac{121}{196}
Add -\frac{6}{7} to \frac{289}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{17}{14}\right)^{2}=\frac{121}{196}
Factor x^{2}+\frac{17}{7}x+\frac{289}{196}. 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{17}{14}\right)^{2}}=\sqrt{\frac{121}{196}}
Take the square root of both sides of the equation.
x+\frac{17}{14}=\frac{11}{14} x+\frac{17}{14}=-\frac{11}{14}
Simplify.
x=-\frac{3}{7} x=-2
Subtract \frac{17}{14} from both sides of the equation.
x ^ 2 +\frac{17}{7}x +\frac{6}{7} = 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 7
r + s = -\frac{17}{7} rs = \frac{6}{7}
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{17}{14} - u s = -\frac{17}{14} + u
Two numbers r and s sum up to -\frac{17}{7} exactly when the average of the two numbers is \frac{1}{2}*-\frac{17}{7} = -\frac{17}{14}. 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{17}{14} - u) (-\frac{17}{14} + u) = \frac{6}{7}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{6}{7}
\frac{289}{196} - u^2 = \frac{6}{7}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{6}{7}-\frac{289}{196} = -\frac{121}{196}
Simplify the expression by subtracting \frac{289}{196} on both sides
u^2 = \frac{121}{196} u = \pm\sqrt{\frac{121}{196}} = \pm \frac{11}{14}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{17}{14} - \frac{11}{14} = -2.000 s = -\frac{17}{14} + \frac{11}{14} = -0.429
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.