Answer
No solution.
Work Step by Step
$3a^2+4a+3=0$
$x=(-b±\sqrt {b^2-4ac})/2a$
$x=(-4±\sqrt {4^2-4*3*3})/2*3$
$x=(-4±\sqrt {16-36})/6$
$x=(-4±\sqrt {-20})/6$
We can't have the square root of a negative number.
Algebra 1
by Hall, Prentice
Published by
Prentice Hall
ISBN 10:
0133500403
ISBN 13:
978-0-13350-040-0
Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Mixed Review - Page 625: 60
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