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2x^2+9x-5=2x-10
We move all terms to the left:
2x^2+9x-5-(2x-10)=0
We get rid of parentheses
2x^2+9x-2x+10-5=0
We add all the numbers together, and all the variables
2x^2+7x+5=0
a = 2; b = 7; c = +5;
Δ = b2-4ac
Δ = 72-4·2·5
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-3}{2*2}=\frac{-10}{4} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3}{2*2}=\frac{-4}{4} =-1 $
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