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Using the quadratic formula or completing the square would be the way to solve this one.
x=(-b+sqr(b^2-4ac))/2a; (-b-sqrb^2-4ac))/2a
a=2
b=-3
c=-4
x=(3+sqr(9+32))/4
x=(3+sqr41)/4
Once you decide to what accuracy you want to get the square root of 41 then you can get the decimal equivalents. A little more than negative 1 and a little more than positive two.
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∙ 9y ago
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What is the nature of the roots of the equation 2xsquared-6x plus 2 equals 0?
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In the quadratic equation ax2 plus bx plus c 0 if b2 - 4ac?
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The roots of the equation are [5 +/- sqrt(11)]/2 = 4.158 and 0.842
How do you find the sum of the roots of the equation x2 plus 5x plus 9 equals 0?
This quadratic equation has no real roots because its discriminant is less than zero.
Which equation has imaginary roots a.x2-1 equals 0 b.x2-2 equals 0 c.x2 plus x plus 1 equals 0 d.x2-x-1 equals 0?
To find which has imaginary roots, use the discriminant of the quadratic formula (b2 - 4ac) and see if it's less than 0. (The quadratic formula corresponds to general form of a quadratic equation, y = ax2 + bx + c)A) x2 - 1 = 0= 0 - 4(1)(-1) = 4Therefore, the roots are not imaginary.B) x2 - 2 = 0= 0 - 4(1)(-2) = 8Therefore, the roots are not imaginary.C) x2 + x + 1 = 0= 1 - 4(1)(1) = -3Therefore, the roots are imaginary.D) x2 - x - 1 = 0= 1 - 4(1)(-1) = 5Therefore, the roots are not imaginary.The equation x2 + x + 1 = 0 has imaginary roots.
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