## Algebra 1

$3a^2+4a+3=0$ $x=(-b±\sqrt {b^2-4ac})/2a$ $x=(-4±\sqrt {4^2-4*3*3})/2*3$ $x=(-4±\sqrt {16-36})/6$ $x=(-4±\sqrt {-20})/6$ We can't have the square root of a negative number.