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2b^{2}-7b-2-2=0
Subtract 2 from both sides.
2b^{2}-7b-4=0
Subtract 2 from -2 to get -4.
a+b=-7 ab=2\left(-4\right)=-8
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2b^{2}+ab+bb-4. To find a and b, set up a system to be solved.
1,-8 2,-4
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -8.
1-8=-7 2-4=-2
Calculate the sum for each pair.
a=-8 b=1
The solution is the pair that gives sum -7.
\left(2b^{2}-8b\right)+\left(b-4\right)
Rewrite 2b^{2}-7b-4 as \left(2b^{2}-8b\right)+\left(b-4\right).
2b\left(b-4\right)+b-4
Factor out 2b in 2b^{2}-8b.
\left(b-4\right)\left(2b+1\right)
Factor out common term b-4 by using distributive property.
b=4 b=-\frac{1}{2}
To find equation solutions, solve b-4=0 and 2b+1=0.
2b^{2}-7b-2=2
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.
2b^{2}-7b-2-2=2-2
Subtract 2 from both sides of the equation.
2b^{2}-7b-2-2=0
Subtracting 2 from itself leaves 0.
2b^{2}-7b-4=0
Subtract 2 from -2.
b=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-4\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
b=\frac{-\left(-7\right)±\sqrt{49-4\times 2\left(-4\right)}}{2\times 2}
Square -7.
b=\frac{-\left(-7\right)±\sqrt{49-8\left(-4\right)}}{2\times 2}
Multiply -4 times 2.
b=\frac{-\left(-7\right)±\sqrt{49+32}}{2\times 2}
Multiply -8 times -4.
b=\frac{-\left(-7\right)±\sqrt{81}}{2\times 2}
Add 49 to 32.
b=\frac{-\left(-7\right)±9}{2\times 2}
Take the square root of 81.
b=\frac{7±9}{2\times 2}
The opposite of -7 is 7.
b=\frac{7±9}{4}
Multiply 2 times 2.
b=\frac{16}{4}
Now solve the equation b=\frac{7±9}{4} when ± is plus. Add 7 to 9.
b=4
Divide 16 by 4.
b=-\frac{2}{4}
Now solve the equation b=\frac{7±9}{4} when ± is minus. Subtract 9 from 7.
b=-\frac{1}{2}
Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.
b=4 b=-\frac{1}{2}
The equation is now solved.
2b^{2}-7b-2=2
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.
2b^{2}-7b-2-\left(-2\right)=2-\left(-2\right)
Add 2 to both sides of the equation.
2b^{2}-7b=2-\left(-2\right)
Subtracting -2 from itself leaves 0.
2b^{2}-7b=4
Subtract -2 from 2.
\frac{2b^{2}-7b}{2}=\frac{4}{2}
Divide both sides by 2.
b^{2}-\frac{7}{2}b=\frac{4}{2}
Dividing by 2 undoes the multiplication by 2.
b^{2}-\frac{7}{2}b=2
Divide 4 by 2.
b^{2}-\frac{7}{2}b+\left(-\frac{7}{4}\right)^{2}=2+\left(-\frac{7}{4}\right)^{2}
Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
b^{2}-\frac{7}{2}b+\frac{49}{16}=2+\frac{49}{16}
Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.
b^{2}-\frac{7}{2}b+\frac{49}{16}=\frac{81}{16}
Add 2 to \frac{49}{16}.
\left(b-\frac{7}{4}\right)^{2}=\frac{81}{16}
Factor b^{2}-\frac{7}{2}b+\frac{49}{16}. 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(b-\frac{7}{4}\right)^{2}}=\sqrt{\frac{81}{16}}
Take the square root of both sides of the equation.
b-\frac{7}{4}=\frac{9}{4} b-\frac{7}{4}=-\frac{9}{4}
Simplify.
b=4 b=-\frac{1}{2}
Add \frac{7}{4} to both sides of the equation.