Answered By Swapan S

5x^2+2x+15=4x^2+13x-15

=> X^2-11x+30=0

Two real roots

$\times=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$×‍=−b±√b²−4ac2a  

Solving for X we get:

X=5,6

Answered By Swapan S

 $5x^2+2x+15=4x^2+13x-15$5x²+2x+15=4x²+13x−15 

=>  $x^2-11x+30=0$x²−11x+30=0 

=>   $x^2-5x-6x+30=0$x²−5x−6x+30=0  

=>   $x\left(x-5\right)-6\left(x-5\right)=0$x(x−5)−6(x−5)=0 

=> $\left(x-5\right)\left(x-6\right)=0$(x−5)(x−6)=0 

=>  $\left(x-5\right)=0;\left(x-6\right)=0$(x−5)=0;(x−6)=0 

=>  $x=5,6$x=5,6  

Answered By Swapan S

See graph

Answered By Emily D

 $5x^2+2x+15=4x^2+13x-15$5x²+2x+15=4x²+13x−15 

The first step, regardless of how we solve this, is to get all the terms onto one side of the equation.

  $5x^2-4x^2+2x-13x+15+15=0$5x²−4x²+2x−13x+15+15=0  

  $x^2-11x+30=0$x²−11x+30=0  

Roots: product of 30, sum of -11, two numbers that fit this description is [-5, -6]

 $\left(x-5\right)\left(x-6\right)=0$(x−5)(x−6)=0 

 $x-5=0$x−5=0 , x = 5

 $x-6=0$x−6=0 , x = 6

Quadratic Formula: That big old fun thing

 $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$x=−b±√b²−4ac2a  , b = -11, a = 1, c = 30

 $x=\frac{-\left(-11\right)+\sqrt{\left(-11\right)^2-4\left(1\right)\left(30\right)}}{2\left(1\right)}$x=−(−11)+√(−11)²−4(1)(30)2(1)  = 6

 $x=\frac{-\left(-11\right)-\sqrt{\left(-11\right)^2-4\left(1\right)\left(30\right)}}{2\left(1\right)}$x=−(−11)−√(−11)²−4(1)(30)2(1)  = 5

Graph:

The first intersect occurs at (5, 150), the second occurs at (6, 207), if you have a Ti calculator, you can search for the intersects on it. Not sure if this is the third way your instructor talked about it.