True or false that integers can be negative numbers
1 answer:
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Answer:
2 real solutions
Step-by-step explanation:
Remember this messy thing?
The <em>quadratic formula</em>, as it's called, gives us the roots to any quadratic equation in standard form (ax² + bx + c = 0). The information on the <em>type</em> of roots is contained entirely in that bit under the square root symbol (b² - 4ac), called the <em>discriminant</em>. If it's non-negative, we'll have <em>real</em> roots, if it's negative, we'll have <em>complex roots</em>.
For our equation, we have a discrimant of (-3)² - 4(6)(-4) = 9 + 96 = 105, which is non-negative, so we'll have real solutions, and since quadratics are degree 2, we'll have exactly 2 real solutions.
