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Find the x intercept of the parabola y=2x^2+6x-5
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No you...I insist
there are two x-ints and the poly does not factor; you can use the quadratic formula or complete the square. divide through by 2 to complete the square\[x^2+3x+\left( \frac{3}{2} \right)^2=\frac{5}{2}+\frac{9}{4}\] factor the LHS; simplify the RHS\[(x+3/2)^2=\frac{19}{4}\]take square root both sides and subtract 3/2\[x=-\frac{3}{2} \pm \frac{\sqrt{19}}{2}\]these are the x-ints
the graph will cross the x-axis at the points\[\left( -\frac{3}{2}, \pm \frac{\sqrt{19}}{2} \right)\]
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