how many real-number solutions does the equation have? –7x² + 6x + 3 = 0

use the quadratic formula to solve the quadratics

b^2 - 4ac = (6)^2 - 4(-7)(3) = 36 + 84 = 120 since b^2 - 4ac> 0 i.e. b^2 - 4ac is positive so the given quadratic equation will have two real roots

Harkirat is using the discriminant to determine the number of solutions. It still is the 3 possible solutions that I mentioned earlier 1 solution 2 solutions No solution

@precal pls do not confuse her...........my method is the right one to determine how many real solutions does a given quadratic equation have

Not confusing anyone just stating the method \[b^2-4ac\] is officially the discriminant

@ Harkirat If the discriminant is greater than 0 then you have 2 real solutions If the discriminant is equal to zero then you have one real solution (also known as a double root) If the discriminant is less than 0 then you have two imaginary solutions (also known as 2 complex solutions) What does this all mean, It means how many times the function crosses the x axis. two solutions (function crosses 2 places or has 2 x intercepts) one solution (function crosses 1 place or has one x intercept) no solution (function does not cross the x axis or no x intercept)

thanks guys!!!

anytime :)

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