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a+b=7 ab=2\left(-15\right)=-30
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-15. To find a and b, set up a system to be solved.
-1,30 -2,15 -3,10 -5,6
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 -30.
-1+30=29 -2+15=13 -3+10=7 -5+6=1
Calculate the sum for each pair.
a=-3 b=10
The solution is the pair that gives sum 7.
\left(2x^{2}-3x\right)+\left(10x-15\right)
Rewrite 2x^{2}+7x-15 as \left(2x^{2}-3x\right)+\left(10x-15\right).
x\left(2x-3\right)+5\left(2x-3\right)
Factor out x in the first and 5 in the second group.
\left(2x-3\right)\left(x+5\right)
Factor out common term 2x-3 by using distributive property.
x=\frac{3}{2} x=-5
To find equation solutions, solve 2x-3=0 and x+5=0.
2x^{2}+7x-15=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{-7±\sqrt{7^{2}-4\times 2\left(-15\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 7 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-7±\sqrt{49-4\times 2\left(-15\right)}}{2\times 2}
Square 7.
x=\frac{-7±\sqrt{49-8\left(-15\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-7±\sqrt{49+120}}{2\times 2}
Multiply -8 times -15.
x=\frac{-7±\sqrt{169}}{2\times 2}
Add 49 to 120.
x=\frac{-7±13}{2\times 2}
Take the square root of 169.
x=\frac{-7±13}{4}
Multiply 2 times 2.
x=\frac{6}{4}
Now solve the equation x=\frac{-7±13}{4} when ± is plus. Add -7 to 13.
x=\frac{3}{2}
Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.
x=-\frac{20}{4}
Now solve the equation x=\frac{-7±13}{4} when ± is minus. Subtract 13 from -7.
x=-5
Divide -20 by 4.
x=\frac{3}{2} x=-5
The equation is now solved.
2x^{2}+7x-15=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.
2x^{2}+7x-15-\left(-15\right)=-\left(-15\right)
Add 15 to both sides of the equation.
2x^{2}+7x=-\left(-15\right)
Subtracting -15 from itself leaves 0.
2x^{2}+7x=15
Subtract -15 from 0.
\frac{2x^{2}+7x}{2}=\frac{15}{2}
Divide both sides by 2.
x^{2}+\frac{7}{2}x=\frac{15}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+\frac{7}{2}x+\left(\frac{7}{4}\right)^{2}=\frac{15}{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.
x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{15}{2}+\frac{49}{16}
Square \frac{7}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{169}{16}
Add \frac{15}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{7}{4}\right)^{2}=\frac{169}{16}
Factor x^{2}+\frac{7}{2}x+\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(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{169}{16}}
Take the square root of both sides of the equation.
x+\frac{7}{4}=\frac{13}{4} x+\frac{7}{4}=-\frac{13}{4}
Simplify.
x=\frac{3}{2} x=-5
Subtract \frac{7}{4} from both sides of the equation.