Question

Solve the equation 2x^2 + 16x + 21 = 0 to the nearest tenth.

Answer 1

Answer:

Rounded to the nearest tenth the solutions to the equation are:

x = -1.7   and x = -6.3

Step-by-step explanation:

Since this is a quadratic equation of the form:

$ax^2+bx+c=0$

Use the quadratic formula, which tells you that the solutions would be given by:

$x=\frac{-b+-\sqrt{b^2-4ac} }{2a}$

So in our case, the quadratic formula gives:

$x=\frac{-b+-\sqrt{b^2-4ac} }{2a}\\x=\frac{-16+-\sqrt{16^2-4(2)(21)} }{2*2}\\x=\frac{-16+-\sqrt{88} }{4}$

which gives us two real solutions:

x = - 1.654792

x = -6.345207

Which rounded to the nearest tenth give:

x = -1.7   and x = -6.3