Find discriminant of quadratic equation:
$$D=(-2)^2-4\cdot(-1)\cdot15=4+60=64$$
Discriminant is greater than zero, so there are two (unequal) real solutions of the equation:
$$x_{1,2}=\dfrac{2\pm8}{2\cdot(-1)}={3,-5}$$

Solution of quadratic equations.
A quadratic equation is any equation having the form \(ax^2+bx+c=0\) where \(x\) represents an unknown, and \(a,b\) and \(c\) represent known numbers such that \(a\) is not equal to 0.
There are two stages in solving quadratic equations.
1) Find discriminant using formula \(D=b^2-4ac\)
2) Check the sign of discriminant:
2.1) If \(D>0\), then equation has two (unequal) real solutions \(x_{1,2}=\dfrac{-b\pm\sqrt{D}}{2\cdot a}\)
2.2) If \(D=0\), then equation has two (equal) real solutions \(x=\dfrac{-b}{2\cdot a}\)
2.3) If \(D<0\), then equation has no real solutions.

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