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6x^2+16x-3=2x^2+5x=0
We move all terms to the left:
6x^2+16x-3-(2x^2+5x)=0
We get rid of parentheses
6x^2-2x^2+16x-5x-3=0
We add all the numbers together, and all the variables
4x^2+11x-3=0
a = 4; b = 11; c = -3;
Δ = b2-4ac
Δ = 112-4·4·(-3)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*4}=\frac{-24}{8} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*4}=\frac{2}{8} =1/4 $
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