One x-intercept for a parabola is at the point(-0.33,0). Use the quadratic formula to find theother x-intercept for the parabola defined bythis equationy=-3x2 + 5x + 2

Answer:(2,0)Step-by-step explanation:Given:The equation is given as: [tex]y=-3x^{2}+5x+2[/tex]For x intercept, [tex]y=0[/tex].Therefore, [tex]-3x^{2}+5x+2=0[/tex]Now, comparing this with the standard quadratic equation [tex]ax^{2}+bx+c=0[/tex], we get[tex]a=-3,b=5,c=2[/tex]Now, using quadratic formula for the above equation, [tex]x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\\x=\frac{-5 \pm \sqrt{5^{2}-4(-3)(2)}}{2(-3)}\\x=\frac{-5 \pm \sqrt{25+29}}{-6}\\x=\frac{-5 \pm \sqrt{49}}{-6}\\x=\frac{-5 \pm 7}{-6}\\x=\frac{-5-7}{-6}\textrm{ or }x=\frac{-5+7}{-6}\\x=\frac{-12}{-6}\textrm{ or }x=\frac{2}{-6}\\x=2\textrm{ or }x=-\frac{1}{3}=-0.33[/tex]Therefore, there are two x intercepts. One was given as (-0.33,0). So, the other one is (2, 0).